I think this can be case bashed, three major cases when $x=max(x,y,z)$ and the other two. Then we can use some weak bounds and induction to bound $x,y,z$.

We have given the Diophantine equation $(1) \;\; xyz = x! + y^x + y^z + z!$, where $x,y,z$ are natural numbers. We may (since $x$ and $z$ are symmetric in (1)) WLOG assume $x \leq z$. Hence, if $z=1$, then $x=1$, yielding $y = 2y + 2$ by (1), i.e. $y=-2 \not \in \mathbb{N}$,. Thus we have $z \geq 2$. According to (1) $z^2y \geq xyz > y^z + y$ i.e. $(2) \;\; z^2 > y^{z-1} + 1$. If $y \geq 3$, then $z^2 > 3^{z-1} + 1$ by (2), which is impossible. Consequently $y \leq 2$. Assume $y=1$. Then by (1) $xz = x! + z! + 2 > x + z(z - 1)$, yielding $x(z - 1) > z(z - 1)$. which means $x > z$ (since $z \geq 2$). This contradiction implies $y \neq 1$, which again give us $y=2$. Hence by (1) we obtain $(3) \;\; 2xz = x! + z! + 2^x + 2^z$. If $x=1$, then by (3) $2z = 2xz > 2^z + z! \geq 2^z + z$, i.e. $2^z < z$, which is impossible. Therefore $x \geq 2$, which according to (3) give us $2z^2 \geq 2xz = 2^z + z! + 2^x + x! \geq 2^z + z(z - 1) + 2^2 + 2!$, which means $2^z \leq z^2 + z - 6 = (z - 2)(z + 3)$. Hence $z \leq 2$, yielding $z=x=2$ since $z \geq x \geq 2$. Checking we find that $(x,z) = (2,2)$ is not a solution of equation (3). Conclusion: Equation (1) has no solution in natural numbers.