More general: Prove that for equation$$x^{n} + y^{n} = z^{n+1}$$there are infinitely many solutions where $x,y$ and $z$ are different natural numbers,and n it is integer positiv number.

$x=a$, $y=ka$ where $k^n + 1 = a$ gives \[x^n + y^n = a^n + k^na^n = a^n(k^n + 1) = a^{n+1}\]

PRMOisTheHardestExam wrote: $x=a$, $y=ka$ where $k^n + 1 = a$ gives \[x^n + y^n = a^n + k^na^n = a^n(k^n + 1) = a^{n+1}\] This is not done, note that $x=z$ in this case. Iterating a similar argument, we get \begin{align*} x&= u^{n+1}\left(v^n+1\right)\\ y&=u^{n+1}v\left(v^n+1\right)\\ z&=u^n\left(v^n+1\right). \end{align*}as a family satisfying the general version. Now if $u\ne v$ and $u,v>1$ then $x,y,z$ are pairwise distinct.