Disclaimer: The solutions we've shared are just one exciting approach, and there are surely many other wonderful methods out there. We’d love to hear your alternative solutions in the community thread below, so let's keep the creativity flowing!

$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.