If $b = 0$, then according to the equation $2a^2 + b = 0$ from which $a = 0$.

Assume now that $b \neq 0$. As $b^3$ and $(2a^2 + b)^3$ are perfect cubes, their ratio $a$ is the cube of a rational number; as it is an integer, it is the cube of an integer $c$. By taking cubic root from each side of the equation we get a

relation $2c^6 + b = bc$, implying $b(c - 1) = 2c^6$. As $c$ and $c - 1$ are coprime, $c^6$ and $c - 1$ are also coprime. Therefore, $c - 1$ has to divide $2$ and $c$ must be one of $3$, $2$, $0$ or $-1$.

If $c = 3$, then $a = 27$ and we get $2b = 2 \cdot 729$, whence $b = 729$.

If $c = 2$, then $a = 8$ and we get $b = 128$.

If $c = 0$, then we get $-b = 0$, which has already been analysed.

If $c = -1$, then $a = -1$ and $-2b = 2$ from which $b = -1$.