$3^x + 4^y = 5^z$ This is an advanced high-school problem.

Find all natural $x,y$, and $z$ such that 
  $3^x + 4^y = 5^z$.

The only obvious solution I can see is $x=y=z=2$. Are there any other solutions?

First mod the whole thing by 3, then $1^y = 2^z$, so $z = 2z_2$ for some $z_2\in\mathbb{N}$. So $3^x = 5^{2z_2} - 2^{2y} = (5^{z_2} + 2^y)(5^{z_2} - 2^y)$.
If $5^{z_2} - 2^y\ne 1$, then by modding each factor by 3 we find that $z_2\ne y \mod 3$ and $z_2 = y \mod 3$, contradiction.
We can invoke Mihăilescu's theorem to find that $x = y = z = 2$.
