The equation is equivalent to
$$ (x+y)^3 + 20 = 7xy(x+y) $$
From this equation, it is clear that $x+y | 20$. Meanwhile, a slight glance on modulo $7$ gives us $(x+y)^3 \equiv 1\pmod{7}$, so $x + y$ is equivalent to one of $1, 2, 4$ modulo $7$. Looking at all positive divisors of $20$, we find that $x + y \in \{1, 2, 4\}$. Obviously, $x + y > 1$ since $x, y$ are positive integers. If $x + y = 2$, then $x = y = 1$, which does not satisfy the equation. Finally, $x + y = 4$ yields $xy = 3$, so we get $(x, y) = (1, 3)$ or $(x, y) = (3, 1)$.