Since $3$ is the only prime divisor of $x^n+y^n$ ,by Zsigmondy's theorem we have that $3\not | x+y$ or $(x,y)=(2,1),(1,2),n=3,m=2$ Suppoce that $3\not |x+y$.Since $3|x\Leftrightarrow 3|y$ we have that $3\not |x,y$. If $n$ is even then $x^n+y^n\equiv 1+1\equiv2\pmod3$,contradiction If $n$ is odd then $x^n+y^n\equiv x+y\not \equiv 0\pmod3$,contradiction So the only solutions $(x,y,n,m)=(2,1,3,2),(1,2,3,2)$.