Integer solutions of the factorial equation $(x!+1)(y!+1)=(x+y)!$ The problem is: are there solutions for the next equation?
$$(x!+1)(y!+1)=(x+y)!$$
with $x,y\in\mathbb{N}$.
My solution:
$\left(x!+1\right)\cdot \left(y!+1\right) = \left(x+y\right)!$
$x!y!+x!+y!+1= \left(x+y\right)!$
$\displaystyle\frac{x!y!+x!+y!+1}{x!y!}= \displaystyle\frac{\left(x+y\right)!}{x!y!}$
$1+\displaystyle\frac{1}{y!}+\displaystyle\frac{1}{x!}+\displaystyle\frac{1}{x!y!}=\displaystyle\binom{x+y}{x}$
how $1+\displaystyle\frac{1}{y!}+\displaystyle\frac{1}{x!}+\displaystyle\frac{1}{x!y!}\leq{4}$, then $\displaystyle\binom{x+y}{x}\leq{4}$, If  $\displaystyle\binom{x+y}{x}\leq{4}$, is neccesary that $x,y\leq{4}$.
Therefore I can check a possible finite set of solutions $\{(x,y)|x,y\leq{4}\}$.
Is correct my proof?
There are other form?

The pairs $(1,2)$ and $(2,1)$ are the only solutions. To see this assume that $x>1$ and $y>1$. Then $x!$, $y!$ and $(x+y)!$ are even. But $(x!+1)(y!+1)$ is odd, which is a contradiction. Thus, $x=1$ or $y=1$, and since the problem is symmetric in $x$ and $y$ we assume that $y=1$. Then $2(x!+1)=(x+1)!$, which is equivalent to $x!+2=x\cdot x!$. Thus, $x!|2$, and this shows that $x=1$ or $x=2$. But $x=1$ is not possible. Thus, $x=2$.
