jmc

Let $a=x+1$ and $b=y-1.$ Then the equation becomes $$(a+b{)}^2=ab.$$This simplifies to $a^2+ab+b^2=0.$ Completing the square in $a,$ we get $${\left(a+\frac{b}{2}\right)}^2+\frac{3b^2}{4}=0,$$which forces $a=b=0.$ Then $(x,y)=(-1,1),$ so there is only $\boxed{1}$ solution.