Prove that the equation 
$$\begin{equation}
xyz=xy+xz+yz+x+y+z\tag{\*}
\end{equation}$$
has finitely many natural solutions.

My attempt. Suppose that $x \leq y \leq z$ then 
$$
xyz\leq xz+xz+yz+3z \implies xy \leq 2x+y+3 \implies x \leq \frac{y+3}{y-2}
$$ 
Since $$x \leq \frac{y+3}{y-2}=\frac{y-2+5}{y-2}=1+\frac{5}{y-2} \implies (x-1)(y-2) \leq 5,$$
we can conclude that the last inequality has finite number of natural solutions, so there are only finite numbers for $x,y.$
For any fixed $x,y$ the equation (\*) reduces to linear equation for $z$ with finite numbers of solutions.

Is it ok?
Is there any other solutions?