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Answer: $t = 3$ and $(x, y, z)$ is any permutation of $(1, 2, 3)$.
Since the equations are symmetrical with respect to variables $x$, $y$ and $z$, we can assume that $x \le y \le z$. Dividing the second equation by the first one we obtain the equality
$$t + 1 = \frac{(t + 1)!}{t!} = \left(1 + \frac{1}{x}\right) \left(1 + \frac{1}{y}\right) \left(1 + \frac{1}{z}\right). \qquad (1)$$
Assume that $y \ge 3$, then also $z \ge 3$. Now from $x \ge 1$ and (1) we get that $t+1 \le 2 \cdot \frac{4}{3} \cdot \frac{4}{3} < 4$, therefore $t \le 2$, but that contradicts the inequality $y \ge 3$.
Thus $y \le 2$. It leaves us with three possibilities.
* $x = y = 1$. Writing (1) in the form $t + 1 = 4 + \frac{4}{z}$ we conclude that $z \in \{1, 2, 4\}$, but none of these values leads to a solution.
* $x = 1$ and $y = 2$. From (1) we get that
$$t + 1 = 2 \cdot \frac{3}{2} \left(1 + \frac{1}{z}\right) = 3 + \frac{3}{z},$$
what means that $z \in \{1, 3\}$ and as $z \ge y \ge 2$ then $z = 3$ and $t = 3$. One can check, that this is a solution, therefore we get 6 solutions where $t = 3$ and $(x, y, z)$ is any permutation of $(1, 2, 3)$.
* $x = y = 2$. From (1) we get that $t + 1 = \frac{9}{4} + \frac{9}{4z}$. The expression of the right hand side is larger than 2 and less than 3 if $z \ge 4$, what is impossible. Therefore $z \le 3$ and it remains to check that $(x, y, z) = (2, 2, 2)$ and $(x, y, z) = (2, 2, 3)$ are not solutions.