The inequality $(2x - y)^2 \geq 0$ (with equality if and only if $y = 2x$) is equivalent to
$$(2x + y)^2 \geq 8xy.$$
The constraint $(x + 1)(y + 2) = 8$ gives $2x + y = 6 - xy$. Substituting this into the inequality above yields
$$(6 - xy)^2 \geq 8xy,$$
which is equivalent to
$$(xy - 10)^2 \geq 64.$$
As we noted already, equality holds for $y = 2x$. In this case, the constraint becomes $(x + 1)(2x + 2) = 8$ which yields $x = 1$ or $x = -3$ and finally the two pairs $(x, y) = (1, 2)$ and $(x, y) = (-3, -6)$. We easily verify that equality actually holds in both cases.