1) $x = z \Rightarrow 2 \cdot x^{x} = (2x)!$ For $x \geq 2$ we got contradiction by Bertrand-Chebyshev theorem ($(2x)!$ is divisible by $p$ for some prime $p$, satisfying $x < p < 2x$, but $2 \cdot x^{x}$ obviously does not divisible by such $p$). For $x = 1$ we have $z + 1 = (z + 1)!$ So $z = 1$ and $x = z = 1$ is the only solution in this case. 2) Assume WLOG $x > z$. We will prove by strong induction that $(x + z)! > x^{z} + z^{x}$ for all natural $x \geq 2$ and natural $z \leq x - 1$. Base: For $x = 2$ we have the only option for $z$: $z = 1$. But obviously $3! > 3$. Proved. Now let us suppose that the following statement correct for some natural $x$. We will prove that it is also true for $x + 1$. $(x + z + 1)! = (x + z)! \cdot (x + z + 1) > (x^{z} + z^{x}) \cdot (x + z + 1)$ (by assumption). We show that $ (x^{z} + z^{x}) \cdot (x + z + 1) > (x + 1)^{z} + z^{x + 1}$ (1) and the result follows. After some obvious algebraic actions we have $x^{z + 1} + z \cdot x^{z} + x^{z} + x \cdot z^{x} + z^{x} > (x+1)^{z} $ (2). We show that $x^{z + 1} + 1 > (x+1)^{z}$. $(x+1)^{z} = x^{z} + 1 + \sum_{i = 1}^{z-1} C_{z}^i \cdot x^{z-i} < x^{z} + 1 + \sum_{i = 1}^{z-1} z^{i} \cdot x^{z-i} < x^{z} + 1 + \sum_{i = 1}^{z-1} x^{i} \cdot x^{z-i} = z \cdot x^{z} + 1 < x \cdot x^{z} + 1 = x^{z + 1} + 1 $. Proved. Now $x^{z + 1} + z \cdot x^{z} + x^{z} + x \cdot z^{x} + z^{x} \geq x^{z + 1} + 4 > x^{z + 1} + 1 > (x+1)^{z}$ so inequality (2) is true and (1) is true too. So far we got that if general inequality is true for some $x$ and $z \leq x - 1$ then it is true for $x + 1$ and $z \leq x - 1$. To complete our inductive proof we only need to check $z = x$. We will prove that $(2x+1)! > 2(x+1)^{x+1} > x^{x+1} + (x+1)^{x}$ for $x \geq 2$. But it is obviously true because $(2x+1)! = (x+1)\cdot(x+2)\cdot x \cdot \prod_{k=2}^{x} (x+1+k)(x+1-k) > (x + 1) \cdot 2 (x + 1) \cdot (x + 1)^{x -1} = 2(x + 1)^{x + 1} > x^{x+1} + (x+1)^{x}$ so inductive proof is over and we are done.

So $\fbox {x = z = 1}$ is the only solution.