I have a friend who turned $32$ recently.  She has an obsessive compulsive disdain for odd numbers, so I pointed out that being $32$ was pretty good since not only is it even, it also has no odd factors.  That made me realize that $64$ would be an even better age for her, because it's even, has no odd factors, and has no odd digits.  I then wondered how many other powers of $2$ have this property.  The only higher power of $2$ with all even digits that I could find was $2048.$  

So is there a larger power of $2$ with all even digits?  If not, how would you go about proving it?

I tried examining the last $N$ digits of powers of $2$ to look for a cycle in which there was always at least one odd digit in the last $N$ digits of the consecutive powers.  Unfortunately, there were always a very small percentage of powers of $2$ whose last $N$ digits were even.

Edit: Here's a little more info on some things I found while investigating the $N$ digit cycles.

$N$: $2,3,4,5,6,7,8,9$

Cycle length: $20,100,500,2500,12500,62520,312500,1562500,\dotsc, 4\cdot 5^{N-1}$

Number of suffixes with all even digits in cycle: $10, 25, 60, 150, 370, 925, 2310,5780,\sim4\cdot2.5^{N-1}$  

It seems there are some interesting regularities there.  Unfortunately, one of the regularities is those occurrences of all even numbers!  In fact, I was able to find a power of $2$ in which the last $33$ digits were even $(2^{3789535319} = \dots 468088628828226888000862880268288)$. 

Yes it's true that it took a power of $2$ with over a billion digits to even get the last $33$ to be even, so it would seem any further powers of $2$ with all even digits are extremely unlikely.  But I'm still curious as to how you might prove it.

Edit 2: Here's another interesting property I noticed.  The next digit to the left of the last $N$ digits will take on every value of its parity as the $N$ digits cycle each time.  Let me illustrate.

The last $2$ digits cycle every $20$ powers.  Now examine the following:

$2^7 =    128$
$2^{27} = \dots 728$
$2^{47} = \dots 328$
$2^{67} = \dots 928$
$2^{87} = \dots 528$
$2^{107} = \dots 128$  

Notice that the hundreds place starts out odd and then proceeds to take on every odd digit as the final 2 digits cycle.

As another example, let's look at the fourth digit (knowing that the last 3 digits cycle every 100 powers.)

$2^{18} = 262144$, 
$2^{118} = \dots 6144$, 
$2^{218} = \dots 0144$, 
$2^{318} = \dots 4144$, 
$2^{418} = \dots 8144$, 
$2^{518} = \dots 2144$  

This explains the power of 5 in the cycle length as each digit must take on all five digits of its parity.

EDIT 3:  It looks like the $(N+1)$st digit takes on all the values $0-9$ as the last $N$ digits complete half a cycle.  For instance, the last $2$ digits cycle every $20$ powers, so look at the third digit every $10$ powers:

$2^{8} = 256$, 
$2^{18} = \dots 144$, 
$2^{28} = \dots 456$, 
$2^{38} = \dots 944$, 
$2^{48} = \dots 656$, 
$2^{58} = \dots 744$, 
$2^{68} = \dots 856$, 
$2^{78} = \dots 544$, 
$2^{88} = \dots 056$, 
$2^{98} = \dots 344$  

Not only does the third digit take on every value 0-9, but it also alternates between odd and even every time (as the Edit 2 note would require.)  Also, the N digits cycle between two values, and each of the N digits besides the last one alternates between odd and even.  I'll make this more clear with one more example which looks at the fifth digit:

$2^{20} = \dots 48576$, 
$2^{270} = \dots 11424$, 
$2^{520} = \dots 28576$, 
$2^{770} = \dots 31424$, 
$2^{1020} = \dots 08576$, 
$2^{1270} = \dots 51424$, 
$2^{1520} = \dots 88576$, 
$2^{1770} = \dots 71424$, 
$2^{2020} = \dots 68576$, 
$2^{2270} = \dots 91424$

EDIT 4: Here's my next non-rigorous observation.  It appears that as the final N digits cycle 5 times, the $(N+2)$th digit is either odd twice and even three times, or it's odd three times and even twice.  This gives a method for extending an all even suffix.  

If you have an all even N digit suffix of $2^a$, and the (N+1)th digit is odd, then one of the following will have the (N+1)th digit even:

$2^{(a+1*4*5^{N-2})}$, 
$2^{(a+2*4*5^{N-2})}$, 
$2^{(a+3*4*5^{N-2})}$

Edit 5: It's looking like there's no way to prove this conjecture solely by examining the last N digits since we can always find an arbitrarily long, all even, N digit sequence.  However, all of the digits are distributed so uniformly through each power of 2 that I would wager that not only does every power of 2 over 2048 have an odd digit, but also, every power of 2 larger than $2^{168}$ has every digit represented in it somewhere.

But for now, let's just focus on the parity of each digit.  Consider the value of the $k^{th}$ digit of $2^n$ (with $a_0$ representing the 1's place.)  

$$
a_k = \left\lfloor\frac{2^n}{10^k}\right\rfloor \text{ mod 10}\Rightarrow a_k = \left\lfloor\frac{2^{n-k}}{5^k}\right\rfloor \text{ mod 10}
$$

We can write 
$$2^{n-k} = d\cdot5^k + r$$
where $d$ is the divisor and $r$ is the remainder of $2^{n-k}/5^k$.  So
$$
a_k \equiv \frac{2^{n-k}-r}{5^k} \equiv d \pmod{10}
$$
$$\Rightarrow a_k \equiv d \pmod{2}$$
And
$$d\cdot5^k = 2^{n-k} - r \Rightarrow d \equiv r \pmod{2}$$
Remember that $r$ is the remainder of $2^{n-k} \text{ div } {5^k}$ so 

$$\text{The parity of $a_k$ is the same as the parity of $2^{n-k}$ mod $5^k$.}$$

Now we just want to show that for any $2^n > 2048$ we can always find a $k$ such that $2^{n-k} \text{ mod }5^k$ is odd.

I'm not sure if this actually helps or if I've just sort of paraphrased the problem.

EDIT 6: Thinking about $2^{n-k}$ mod $5^k$, I realized there's a way to predict some odd digits.  

$$2^a \pmod{5^k} \text{ is even for } 1\le a< log_2 5^k$$

The period of $2^a \pmod{5^k}$ is $4\cdot5^{k-1}$ since 2 is a primitive root mod $5^k$.  Also 

$$2^{2\cdot5^{k-1}} \equiv -1 \pmod{5^k}$$

So multiplying any $2^a$ by $2^{2\cdot5^{k-1}}$ flips its parity mod $5^k$.  Therefore  $2^a \pmod{5^k}\text{ }$ is odd for

$$1 + 2\cdot5^{k-1} \le a< 2\cdot5^{k-1} + log_2 5^k$$

Or taking the period into account, $2^a \pmod{5^k} \text{ }$ is odd for any integer $b\ge0$ such that

$$1 + 2\cdot5^{k-1} (1 + 2b) \le a< 2\cdot5^{k-1} (1 + 2b) + log_2 5^k$$

Now for the $k^{th}$ digit of $2^n$ ($ k=0 \text{ } $ being the 1's digit), we're interested in the parity of $2^{n-k}$ mod $5^k$.  Setting $ a =n-k \text{  } $ we see that the $k^{th}$ digit of $2^n$ is odd for integer $b\ge0$ such that

$$1 + 2\cdot5^{k-1} (1 + 2b) \le n - k < 2\cdot5^{k-1} (1 + 2b) + log_2 5^k$$

To illustrate, here are some guaranteed odd digits for different $2^n$:  

(k=1 digit): $ 2\cdot5^0 + 2 = 4 \le n \le 5 $
(k=2 digit): $ 2\cdot5^1 + 3 = 13 \le n \le 16 $
(k=3 digit): $ 2\cdot5^2 + 4 = 54 \le n \le 59 $
(k=4 digit): $ 2\cdot5^3 + 5 = 255 \le n \le 263 $  

Also note that these would repeat every $4\cdot5^{k-1}$ powers.

These guaranteed odd digits are not dense enough to cover all of the powers, but might this approach be extended somehow to find more odd digits?

Edit 7: The two papers that Zander mentions below make me think that this is probably a pretty hard problem.

This seems to be similar to (I'd venture to say as hard as) a problem of Erdős open since 1979, that the base-3 representation of $2^n$ contains a 2 for all $n>8$.

Here is a paper by Lagarias that addresses the ternary problem, and for the most part I think would generalize to the question at hand (we're also looking for the intersection of iterates of $x\rightarrow 2x$ with a Cantor set). Unfortunately it does not resolve the problem.

But Conjecture 2' (from Furstenberg 1970) in the linked paper suggests a stronger result, that every $2^n$ for $n$ large enough will have a 1 in the decimal representation. Though it doesn't quantify "large enough" (so even if proved wouldn't promise that 2048 is the largest all-even decimal), it looks like it might be true for all $n>91$ (I checked up to $n=10^6$).
