The trivial solutions are $(\pm 1,\pm 1)$. Let $d = \gcd(x+1,y+1)$ and $x+1 = dx_0$ and $y+1 = dy_0$ with $\gcd(x_0,y_0) = 1$, we have $(x_0(x-1) + y_0)(y_0(y-1) + x_0) = x_0y_0 \to x_0y_0(x-1)(y-1) + y_0^2(y-1) + x_0^2(x-1) = 0$. The determinant is $\Delta_{x_0} = -3y_0^2(x-1)^2(y-1)^2 \leq 0$. Therefore, $y_0 = 0$, the case $x=1$ and $y=1$ are the previously mentioned trivial solutions. Analogously, we have $x_0 = y_0 = 0 \to x=y=-1$. The only solutions are the trival. $$(1,1),(1,-1),(-1,1),(-1,-1)$$Q.E.D.