Through pure speculation, we can see that a solution satisfying $ (m,n)=(4,1)$. Now we will show that this is our only solution to the case.
Suppose that without loss of generality, we have positive integers $ a,b$ s.t. $ a>b>1$.
Then we can see that
$ \lim_{b\to\infty}b^a-a^b=\infty>3$
Once it reaches a point, it just becomes greater than $ 3$.
If we consider a different expression, $ a^b-b^a$, we can see that this will fail as well for $ a>b>1$:
Similarly, the expression will not be satisfied to equal 3 for small positive integers and as $ b\to\infty$
$ \lim_{b\to\infty}a^b-b^a=-\infty$
Thus, our only solution will be $ m=4$ and $ n=1$