  Per the $ABC$ conjecture, for every $\varepsilon > 0$, there exist finitely
many triples $\{A, B, C\}$, satisfying $A + B = C, gcd(A,B) = 1, B > A \geq 1$,
such that $C > rad(ABC)^{1 + \varepsilon}$, where $rad(ABC)$ is the product of
all distinct primes constituting $ABC$. It is shown (possibly known) that for
any $\varepsilon > 0$, $rad(ABC)^{1 + \varepsilon} > C$ holds for $ABC$ of the
form $2^mp^n$ such that $2^m + \mu = p^n$, where $m, n$ are positive integers,
$\mu = \pm 1$, and $p$ is an odd prime. Condition $2^m + \mu = p^n$ requires
$p$ to be either a Mersenne or Fermat prime. For $ABC = 2^mp^nq^r$, $p$ and $q$
either Fermat or Mersenne primes but distinct, and $m,n,r$ positive integers,
such that any diophantine relation, $\mbox{(i)}\; 2^m + \mu = p^nq^r,
\mbox{(ii)} \; 2^mp^n + \mu = q^r \mbox{ or } \mbox{(iii)} \; p^n + \mu q^r =
2^m$, holds, using results in literature and mostly elementary methods, the
number of triples $\{A, B, C \}$ are shown to be finite, except for $\{2^{y+1},
2^{2y}+1, (2^y+1)^2\}$, a solution to (iii) with $y$ such that $2^y+1$ and
$2^{2y}+1$ are primes, in which case $rad(ABC)^{1 + \varepsilon} > C$ holds for
any $\varepsilon > 0$.
