TITLE: Ordered pairs of Integers $(x,y)$ which satisfy $x!\cdot y! = x!+y!+2$
QUESTION [7 upvotes]: Find the total number of ordered pairs $(x,y)\in\mathbb{Z}^2$ which satisfy
$$
\tag1 x!y! = x!+y!+2
$$

My Attempt:
We can write $(1)$ as 
$$
\begin{align}
x!y!-x!-y!+1 &= 3\\
\left(x!-1\right)\left(y!-1\right) &= 3=1\times 3 = 3\times 1
\end{align}
$$
So, without loss of generality, $(x!-1)=1$ and $(y!-1)=3$, which implies that  $x!=2$ and $y!=4$. But there is no $y\in\mathbb{Z}$ such that $y!=4$. Thus no ordered pairs $(x,y)\in\mathbb{Z^2}$ exist which satisfy the given equation.
Is this solution correct? If not, why not?

REPLY [2 votes]: Here is another proof: 
Assume $x\geq y\geq 3$. Then we have $$x!\,y!\geq x!\,3!>3\cdot x!=x!+x!+x!>x!+y!+2$$ The cases $0<x,y<3$ are checked by hand.