We can rewrite the equation as follows:

$$(x+y)\left(x^{2}-x y+y^{2}\right)=4 x y(x+y)-20$$

Now, $x+y$ is a divisor of the left side and of the first term on the right, so it must also divide the second term on the right: $x+y \mid 20$. Since $x+y \geq 2$, this gives the possibilities for $x+y$ as $2,4,5,10,20$. If exactly one of $x$ and $y$ is even and the other is odd, the left side of the equation is odd and the right side is even, which is a contradiction. Therefore, $x+y$ must be even, ruling out $x+y=5$. If $x+y=2$, then $x=y=1$, and the left side is positive while the right side is negative, so this possibility is also ruled out.
To try the other possibilities, we rewrite the equation slightly differently:

$$(x+y)\left((x+y)^{2}-3 x y\right)=4 x y(x+y)-20$$

If $x+y=4$, we have $4 \cdot(16-3 x y)=16 x y-20$, so $16-3 x y=4 x y-5$, thus $21=7 x y$, or $x y=3$. Therefore, $(x, y)=(3,1)$ or $(x, y)=(1,3)$. Both pairs satisfy the equation.
If $x+y=10$, we get $100-3 x y=4 x y-2$, so $7 x y=102$. But 102 is not divisible by 7, so this is not possible.
If $x+y=20$, we get $400-3 x y=4 x y-1$, so $7 x y=401$. But 401 is not divisible by 7, so this is not possible.
Having exhausted all possibilities, we conclude that $(1,3)$ and $(3,1)$ are the only solutions.