$3^x + 4^y = 5^z$ 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?

Note 
$$2^z\equiv (3+2)^z=5^z=3^x+4^y\equiv 0+1=1\pmod 3$$
so
$z$ is even number,let $z=2z_{1}$,then we have
$$(5^{z_{1}}+2^y)(5^{z_{1}}-2^y)=3^x$$
so we have
$$5^{z_{1}}+2^y=3^x,5^{z_{1}}-2^y=1$$
then we have
$$(-1)^{z_{1}}+(-1)^y\equiv 0\pmod 3,(-1)^{z_{1}}-(-1)^y\equiv 1\pmod 3$$
so we have $z_{1}$ is odd,$y$ is even,let $y=2y_{1}$
we have
$$(4+1)^{z_{1}}+4^{y_{1}}=(4-1)^x\Longrightarrow (-1)^x\equiv 1\pmod 4$$
so $x$ is also even.
if $y>2$, since
$$5^{z_{1}}+2^y=3^x\Longrightarrow 5\equiv 1\pmod 8$$
that's not impossible.so
$y=2\Longrightarrow z_{1}=1,z=2,x=2$$
