Simon's Favorite Factoring Trick Problem Suppose that $x,y,z$ are positive integers satisfying $x \le y \le z$, and such that the product of all three numbers is twice their sum. What is the sum of all possible values of $z$?
Since this is for only positive integers, and there are sums and products involved, I think that this can be approached using Simon's Favorite Factoring Trick. I am not sure how though. Help is greatly appreciated
For those who do not know what Simon's Favorite Factoring Trick is, it is a method of factoring by grouping. For example, say that we want to factor $xy+x+y+1$. We can factor this as so:
$$xy+x+y+1$$
$$x(y+1)+y+1$$
$$x(y+1)+1(y+1)$$
$$(x+1)(y+1)$$

Suppose $x,y,z$ are positive integers, with $x \le y \le z$, such that $xyz=2(x+y+z)$.
\begin{align*}
\text{Then}\;\;&xyz=2(x+y+z)\\[4pt]
\implies\;&xyz \le 2(3z)\\[4pt]
\implies\;&xy \le 6\\[4pt]
\implies\;&x^2 \le 6\\[4pt]
\implies\;&x \le 2\\[4pt]
\end{align*}
Consider two cases . . .

Case $(1)$:$\;x=1$.
\begin{align*}
\text{Then}\;\;&xyz=2(x+y+z)\\[4pt]
\iff\;&yz =2(1+y+z)\\[4pt]
\iff\;&yz - 2y -2z  - 2 = 0\\[4pt]
\iff\;&(y-2)(z-2)=6
\end{align*}
which leads to a small number of possibilities for $y,z$, left for you to complete.

Case $(2)$:$\;x=2$.

Since $xy\le 6$, and $x \le y$, we get $2\le y\le 3$.

Using $x=2$, for each of $y=2,y=3$, solve the equation $xyz=2(x+y+z)$ for $z$.
