Obviously it's enough to prove only one direction of the assertion. Let us set $x = a c^{u}\,,\, y=b c^{u}\,,\, z= c^v$. Plugging it into $x^n+y^n = z^k$, we have: \[ (a^n+b^n) = c^{vk-nu}\]Choose now $u,v$ such that $vk-nu=n-k$ and we obtain \[ a^n+b^n=c^{n-k}\,\,\,\,\,\,\, (1) \]Hence, if $a,b,c$ satisfy (1), then the corresponding $x,y,z$ will satisfy $x^n+y^n = z^k$. Comment: I've seen a similar idea used to prove $x^n+y^n=z^m$ has infinity many solutions when $m,n$ are relatively prime.