$2^n+3^n= m^k$ imply $k=1$

Question

If $m,n,k$ are natural numbers such that $$2^n+3^n= m^k$$ where demonstrate that $k=1.$

I tried to prove that divides equal amount left 5 but does not divide by 25 and we did. Maybe someone else has the idea. Thank you!

Answer 1

For $k=2$ this has been shown, among other papers, in More on a Diophantine equation $2^x+3^y=m^2$. It has only three solutions, namely $(x,y)=(0,1),(3,0),(4,2)$. In no case we have $x=y$. Hence we solved the question for all even $k$. For $k$ odd, it follows that $n=k$ (I cannot find the reference right now, but I will try). Then, by Fermat, this is impossible.