Answer: $a = b = t$, $t \in \mathbb{Z}$, $(a; b) = (-18; -2)$, $(a; b) = (0; 7)$, $(a; b) = (12; 3)$.

We rearrange the initial equation $(b^2 + 7(a-b))^2 = a^3 b$:
$$\begin{align*} (b^2 + 7(a-b))^2 &= a^3 b \Leftrightarrow b^4 + 14b^2(a-b) + 49(a-b)^2 = a^3 b \Leftrightarrow \\ &\qquad a^3 b - b^4 - 14b^2(a-b) - 49(a-b)^2 = 0 \Leftrightarrow \\ &\qquad b(a^3 - b^3) - 14b^2(a-b) - 49(a-b)^2 = 0 \Leftrightarrow \\ &\qquad (a-b)(ba^2 + ab^2 + b^3 - 14b^2 - 49(a-b)) = 0. \tag{1} \end{align*}$$
If $a = b$, then it is evident that (1) holds. Therefore we have the infinite set of the integer solutions: $a = b = t$, where $t \in \mathbb{Z}$.

Let $a \neq b$, then from (1) it follows that
$$ba^2 + ab^2 + b^3 - 14b^2 - 49(a-b) = 0. \quad (2)$$
If $b=0$, then from (2) it follows that $a=0$, i.e. $a=b=0$, a contradiction. Thus $b \neq 0$. Consider (2) as a quadratic equation with respect to $a$.
$$ba^2 + (b^2 - 49)a + b^3 - 14b^2 + 49b = 0. \quad (3)$$
The discriminant $D(b)$ of this equation is
$$\begin{align*} D(b) = (b^2 - 49)^2 - 4b(b^3 - 14b^2 + 49b) &= ((b-7)(b+7))^2 - 4b^2(b-7)^2 = \\ &= (b-7)^2((b+7)^2 - 4b^2) = -(b-7)^3(3b+7). \end{align*}$$
Equation (3) has real solutions iff $D(b) \ge 0$, i.e. for $b \in [-7/3; 7]$. The segment $[-7/3; 7]$ contains exactly ten integers: $-2, -1, 0, 1, 2, 3, 4, 5, 6$ and $7$. Since $b \neq 0$ and $a_{1,2} = (49 - b^2 \pm \sqrt{D})/(2b)$, we have exactly four integers ($b = -2, 3, 6, 7$) such that $D(b)$ is a perfect square (we need integer solutions of (3)).

1) If $b = -2$, then $D = 27^2$ and either $a_1 = -9/2$ or $a_2 = -18$, so, in this case $(a; b) = (-18; -2)$ is a required solution.

2) If $b = 3$, then $D = 2^{10}$ and either $a_1 = 4/3$ or $a_2 = 12$, so $(a; b) = (12; 3)$ is a required solution.

3) If $b = 6$, then $D = 5^2$ and either $a_1 = 13/12$ or $a_2 = 85/12$, so there are no solutions in this case.

4) If $b = 7$, then $D = 0$ and $a_{1,2} = 0$, so $(a; b) = (0; 7)$.