Answer: $(\frac{z^3(z-1)}{2}, \frac{z^6(z-1)}{2})$ where $z$ is arbitrary integer.

Solution:
If $x = 0$ then $y = 0$. We now assume that $x \neq 0$.
Let $d = \gcd(x, y) > 0$ and $x = da, y = db$. Dividing the sides of the equation by $d^4$, we get $b^4 = a(2a^2d + b)^3$. Thus $a \mid b^4$. Since $a$ and $b$ are coprime, the only possibility is $|a| = 1$ and the equation reduces to
$$b^4 = \pm (2d + b)^3.$$
As the same integer is simultaneously a cube and a fourth power, it is a twelfth power of some integer, yielding $b = z^3$ for some integer $z$. Substituting $b = z^3$ into the equation and taking cube root gives $z^4 = \pm(2d + z^3)$ or, equivalently, $\pm z^4 = 2d + z^3$. Since $-z^4 \le -|z^3| \le z^3 < 2d + z^3$, the minus sign is not possible, therefore $d = \frac{z^4-z^3}{2} = \frac{z^3(z-1)}{2}$. We conclude that $x = \frac{z^3(z-1)}{2}$ and $y = \frac{z^6(z-1)}{2}$, where $z$ is an arbitrary integer different from $0$ and $1$. In this case $d > 0$ and the solution satisfies the original equation. If $z = 0$ or $z = 1$, we get the initial solution $(0, 0)$.