Since $xy^2 + 2y$ divides
$$(2x + y)(xy^2 + 2y) - y(2x^2y + xy^2 + 8x) = 2y^2 - 4xy,$$
we conclude that $xy + 2$ divides $2y - 4x$. We consider two cases.

Case 1. Let $2y - 4x \ge 0$. Then we have two possibilities:

1.1) If $x \ge 2$ then $xy + 2 > 2y - 4x$. Hence $2y - 4x = 0$, i.e. $x = a$ and $y = 2a$, where $a \ge 2$ is an integer. Since $xy^2 + 2y = 4a(a^2 + 1)$ divides $2x^2y + xy^2 + 8x = 8a(a^2 + 1)$, this gives a solution.

1.2) If $x = 1$ then $y^2 + 2y$ divides $8$, i.e. $y = 2$.

Hence in this case the solutions are $(a, 2a)$, where $a$ is a positive integer.

Case 2. Let $2y - 4x < 0$, i.e. $4x - 2y > 0$. If $y \ge 4$, then $xy + 2 > 4x - 2$. Therefore $y = 1, 2$ or $3$.

2.1) If $y = 1$ then $\frac{2x^2 + 9x}{x + 2} = 2x + 5 - \frac{10}{x + 2}$ is an integer. Hence $x + 2$ divides $10$ and this gives the solutions $x = 3, y = 1$ and $x = 8, y = 1$.

2.2) If $y = 2$ then $\frac{x^2 + 3x}{x + 1} = x + 2 - \frac{2}{x + 1}$ is an integer which gives $x = 1$.

2.3) If $y = 3$ then $\frac{6x^2 + 17x}{9x + 6}$ is an integer. This implies that $3|x$, i.e. $x = 3k$ for some positive integer $k$. After simplifications we obtain that $\frac{18k^2 + 17k}{9k + 2} = (2k + 1) + \frac{4k - 2}{9k + 2}$ is an integer, which is impossible for $k \ge 1$.

Finally, the solutions are $x = a, y = 2a$ for all positive integers $a$ and $x = 3, y = 1, x = 8, y = 1$.