Prove that the equation $ xyz=xy+xz+yz+x+y+z$ has finite numbers of natural solutions. 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?

Add $xyz+1$ to both sides to obtain $$2xyz+1=xyz+xy+yz+zx+x+y+z+1=(x+1)(y+1)(z+1)$$
The left-hand side is odd, so the right-hand side must also be odd, whence $x,y,z$ are all even.
Now divide by $xyz$ to obtain $$\left(1+\frac 1x\right)\left(1+\frac 1y\right)\left(1+\frac 1z\right)=2+\frac 1{xyz}\gt 2$$
Now note that $\left(\frac 54\right)^3=\frac {125}{64}\lt 2$, so that one of $x,y,z$ at least must be equal to $2$ - say $x$.
We then have $$\frac 32\left(1+\frac 1y\right)\left(1+\frac 1z\right)=2+\frac 1{2yz}$$ which gives $$\left(1+\frac 1y\right)\left(1+\frac 1z\right)\gt \frac 43$$ from which it follows that one of $x$ and $y$ must be less than $8$. Note that it can't be $6$, because that would not work in the top equation of this answer taken mod $3$. Once $y$ is fixed, there is at most one possibility for $z$ (the equation is linear in $z$) and any answer must be a permutation of one with $x\le y\le z$.
Using this method, you should be able to find the solutions.
