Diophantus- Diophantine problem

We can asume WLOG that $x≤y<z$ , because if at least one of $x, y$ would be at least $z$ then $z^z<x^x+y^y$ .
So $x^x+y^y≤2(z-1)^{z-1}≤(z-1)(z-1)^{z-1}<z^z$ for every $z\geq3$ .
So we can be sure that for every $z\geq3$ is the following true for every natural numbers $x,y$ : $x^x+y^y\ne z^z$ .
And now there are only few cases that we must check and here they are:
1.) $z=2$
$x^x+y^y=4$ , this equation has no solutions because $max(x,y)$ must be lower than 2 and thus $x=y=1$ which simply does not fulfill the equation.
2.) For $z=1$ this equation has no solutions, because $max(x,y)$ must be lower than 1 and there are no natural numbers lower than 1.
And we are done.