We can easily see that $x, y, z \ge 2$.
If $x, y, z \ge 3$, then observe that $3$ doesn't divide $a, b, c$. Indeed, if for example, $3 \mid a$, then $3$ doesn't divide $ab + 1$, but $3$ divides $x!$, contradiction.
This means that two of the numbers are congruent mod $3$. Due to the symmetry,
suppose that $a \equiv b \pmod{3}$. Then, we have
$$x! = ab + 1 \equiv a^2 + 1 \pmod{3}.$$
However, $3 \mid x!$, so $3 \mid a^2 + 1$, which is absurd, since for all integers $x$, we have
$x^2 \equiv 0,1 \pmod{3}$.
It follows that one of $x, y, z$ is smaller than $3$, let it be $x$. Then $x = 2$ and from
$ab + 1 = 2! = 2$, we get $ab = 1$, so $a = b = 1$. Then, from the second relation,
$y = z$, so all solutions have the form $(1,1, y! - 1)$ with its cyclic permutations.