We have $4=(x+1)(y+1)\le\frac{(x+y+2)^2}4$, so $(x+y+2)^2\ge16\Leftrightarrow x+y+2\ge4\Leftrightarrow x+y=2$. Equality is when $x=y$, so $x=y=1$.

Supose $x+y<2$. Then $1>\frac{x+y}{2}\geq \sqrt{xy}\implies xy<1$ so $x+y+xy<3$, contradiction. When $x+y=2$ we want $y=\frac{1}{x}\implies (x, y)=(1, 1)$