We can assume $x,y$ to be odd, if not we can just reduce them to odd, and it still satisfies. If $n$ is even, then $x^n+y^n\equiv 2 \mod 4 \implies m=1 \implies 1^1+1^1$ is the only possible sol, but $n>1.$ So $n$ is odd, then using LTE, we get, $v_2(x^n+y^n)=v_2(x+y).$ But since $(x^n+y^n)=(x+y)(x^{n-1}+\dots+y^{n-1})=2^m\implies x^{n-1}+\dots+y^{n-1}=1 ,$ but for $n>3.$ Clearly this is false. When $n=1$, just consider $(x,y)=(1,1).$ Then we can just multiply powers of $2$ on each side, so the solutions are $2^k,2^k,2^{k+1} \implies n=k,m=k+1, k>1.$