$x! + 2^y = (x+1)! $

$ \implies 2^y = (x+1)\cdot x! - x!$

$ \implies 2^y = x\cdot x!$

Now, if $x>2$, $3 \mid x\cdot x!$. But $3\nmid 2^y$. $\therefore x \leq 2$. $x=0$ doesn't work because $2^y = 0$ has no solutions.

By putting $x=1,2$, we obtain that $(x,y)=(1,0),(2,2)$ and we are done.