Let $a,b >1$ be integers. When does $a^b \mid b^a$?
Certainly if this is true then $a\mid b$ by considering $a$'s prime factors. (not quite convinced).
Also then if $b$ is prime then $a=b$.
There the well know solution to the problem $x^y=y^x$, which tells that $2^4=4^2$ is the only solution given the conditions.
I've a feeling this might be the only other case.