Prove that $x^x+y^y=z^z$ doesn't have integer solutions
A proof for positive integers.
Suppose $x$ and $y$ are positive integers with $x\le y$. Then $x^x+y^y<2y^y$. We can show that $$ x^x+y^y \le 2y^y < (y+1)^{y+1}. $$ Consider $$f(y) = (y+1)\log(y+1)-y \log y - \log 2.$$ Note $f(1)>0$ and $$f'(y) = \log(y+1) -\log y > 0$$ for all $y>0$. Hence, $f(y)>0$ for all $y>0$ and so $$2y^y < (y+1)^{y+1}$$ for all $y>0$. Thus $x^x+y^y<(y+1)^{y+1}$ and so $x^x+y^y=z^z$ has no solutions in positive integers.
(Note that the other solutions don't deal with the case of negative integers.)
$0^0$ is undefined, so work with these being non-zero. If you believe that $0^0 = 0$ or $1$, that can be easily dealt with, and is left as an exercise to the reader.
We consider cases based on how many of them are positive or negative.
If all of the integers are positive, WLOG $ x \leq y < z$
Show that for $n \geq 1$, $(n+1)^{n+1} > (n+1) n^n \geq 2 n^n$.
Hence $x^x + y^y \leq 2 y^y < (y+1)^{y+1} \leq z^z$, so there are no solutions.
If one of the integers is negative, it cannot be $z$. WLOG let $ x < 0$.
If $x = -1$, clearly there are no solutions.
If $ x < -1$, the the LHS is not an integer, but the RHS is. Hence, no solutions.
If exactly 2 of the integers are negative, then those values of $ |n^n| \leq 1$. The remaining value of $n^n$ from the positive integer is either 1 or $ \geq 4$.
Thus, the only possibility for solutions is $\pm x^x \pm y^y = \pm1$.
If $x = -1$ or $ y = -1$, clearly there are no solutions.
Otherwise, $x, y \leq -2$ and $|n^n| \leq \frac{1}{4}$, and thus this has no solutions.
If all 3 of the integers are negative, if 2 of them are equal, then the third raised to itself is 0, hence no solutions. So all of them are distinct. WLOG, $ x < y < 0$.
If any of them are $-1$, then we can easily rule out solutions (like the previous cases).
Then, for $ n \leq -2$, $\frac{|n^n|} { |(n-1)^{n-1}|} \geq 3 > 2$ in a manner similar to the all-positive case.
As such, we have no solutions to $\pm |x^x| \pm |y^y| = \pm |z^z|$.
Thus, there are no solutions to $x^x + y^y + z^z = 0 $.