Solutions $(x, y),\;x,y\in \mathbb Z$ to the equation $x! +1 = y^2$

Question

Solve $x! +1 = y^2$ over integers.

I don't know what to try here. The equation implies that $x! = (y-1)(y+1)$ but I'm not sure if that helps at all.

Answer 1

The three known solutions $(x,y)$ to Brocard's problem are $(x,y)=(4,5)$, $(x,y)=(5,11)$, and $(x,y)=(7,71)$.