Since the left-hand side is greater than both $x!$ and $y!$, obviously $z > x$ and $z > y$. So, both sides of the equation are divisible by both $x!$ and $y!$. Therefore, $x \cdot y!$ is divisible by $x!$, which means that $y!$ is divisible by $(x-1)!$, giving $y \ge x-1$. Analogously, $2y \cdot x!$ is divisible by $y!$, meaning $2 \cdot x!$ is divisible by $(y-1)!$. The case $x=1, y=3$ is not a solution, the case $x > 1$ gives $2 \cdot x! < (x+1)!$, which implies $x \ge y-1$. This leaves us to look through the cases $-1 \le y-x \le 1$.
* If $y = x - 1$, then the equation simplifies to $(2x-1) \cdot x! = z!$. As $(x+1)(x+2) > 2x-1$, we have $2x-1 = x+1$ and $z = x+1$. This gives the solution $x=2, y=1, z=3$.

* If $y = x$, the equation simplifies to $3x \cdot x! = z!$. As $(x+1)(x+2) > 3x$, we have $3x = x + 1$, but this does not give integer solutions.
* If $y = x + 1$, the equation simplifies to $(x^2 + 3x + 2) \cdot x! = z!$ or $(x + 2)! = z!$. From here we get a family of solutions $x = n, y = n + 1, z = n + 2$.