I will instead characterize every solution $(x,y,m,n)$ given that $x,y>0$ and $m,n \geq 1$—this is the version of the problem I received. Clearly $(x,y,m,n)=(x,y,1,1)$ works, and by size we require $n\geq m$ hence no other solution exists for $n=1$, so assume $n>1$. Then as a consequence of Zsigmondy there must exist some prime dividing $x^n+y^n$ that doesn't divide $x+y$ and thus $(x+y)^m$, unless: $(x,y)=(k,2k)$ for some $k$, and $n=3$ hence $m \in \{1,2\}$. Then either $3k=9k^3$ which is impossible or $9k^2=9k^3 \implies k=1$, which yields $(x,y,m,n)=(1,2,2,3)$ and its permutation $(2,1,2,3)$. $(x,y)=(k,k)$ for some $k$, so $2^mk^m=2k^n$ from which we find that $k$ must be a power of $2$, say $2^j$. Then $$2^{m+mj}=2^{nj+1} \implies j=\frac{m-1}{n-m},$$from which we extract the very ugly answer $\left(2^\frac{m-1}{n-m},2^\frac{m-1}{n-m},m,n\right)$ for all $(m,n)$ such that $n-m \mid m-1$. Since everything is reversible, it follows that the solutions are $(x,y,m,n)=(x,y,1,1)$ $,(1,2,2,3),(2,1,2,3)$, and $\left(2^\frac{m-1}{n-m},2^\frac{m-1}{n-m},m,n\right)$ for all $(m,n)$ such that $n-m \mid m-1$. $\blacksquare$ (clearly, this implies the original problem)