(Belarus MO 2014 Category B Problem 2 modified) The only solutions are $(a, b) = (t, t), (0, 11)$ where $t$ is any integer.
We have
$$\begin{aligned} (b^2 + 11(a-b))^2 &= a^3b \\ \Leftrightarrow \quad b^4 + 22b^2(a-b) + 121(a-b)^2 &= a^3b \\ \Leftrightarrow \quad (a-b)(121(a-b) + 22b^2 - b(a^2 + ab + b^2)) &= 0. \end{aligned}$$
When $a = b$, the equation always holds. Now we may assume $a \neq b$. It remains to solve $121(a-b) + 22b^2 - b(a^2 + ab + b^2) = 0$. This means
$$baa^2 + (b^2 - 121)a + b(b - 11)^2 = 0. \quad (1)$$
If $b=0$, we have $a=0$, which is a contradiction. If $b \neq 0$, this is a quadratic equation in $a$. The discriminant is
$$\Delta = (b^2 - 121)^2 - 4b^2(b - 11)^2 = -(b - 11)^3(3b + 11).$$
When $b=11$, the equation (1) becomes $11a^2 = 0$. The only solution is $a=0$. When $b \neq 11$, we need $-(b-11)(3b+11) \ge 0$. This implies $b = -3, -2, \dots, 10$. Note that $\Delta$ should be a perfect square since $a$ is an integer. We check the value of $c = -(b-11)(3b+11)$ one by one as follows.