We know that $2^4 = 4^2$ and $(-2)^{-4} = (-4)^{-2}$. Is there another pair of integers $x, y$ ($x\neq y$) which satisfies the equality $x^y = y^x$?
This is a classic (and well known problem).

The general solution of $x^y = y^x$ is given by

$$\begin{align*}x &= (1+1/u)^u \\ y &= (1+1/u)^{u+1}\end{align*}$$

It can be shown that if $x$ and $y$ are rational, then $u$ must be an integer.

For more details, see this and this.
