The factorization of the right hand side of the equation gives us
$$y^2 = x(x + 1)(x + 2).$$
Since $x \in \mathbb{Z}$, obtained factors are consecutive integers.
First we assume that all the factors, $x$, $x+1$ and $x+2$ are different from zero. Since the left hand side of the equality is positive, these three factors also have to be positive.
It follows that $x+1$ and $x(x+2)$ are relatively prime (since $x+1$ and $x$ are relatively prime and also $x+1$ and $x+2$ are relatively prime). The product of two relatively prime positive integers is a perfect square if and only if each of the numbers is a perfect square. In particular, that means that
$$x(x + 2) = x^2 + 2x = (x + 1)^2 - 1$$
is a perfect square. Therefore, $(x+1)^2 - 1$ and $(x+1)^2$ are two consecutive integers which are both perfect squares, and that is possible if and only if those numbers are zero and one. Hence,
$$(x+1)^2 = 1$$
so $x = -2$ or $x = 0$. But then $x + 2 = 0$ or $x = 0$, which is a contradiction with the assumption that all the factors are different from zero.
If one of the factors on the right hand side is equal to zero, then $y = 0$ and we get the solutions
$$(x, y) \in \{(-2, 0), (-1, 0), (0, 0)\}.$$