Find all integer pairs $(x, y)$ such that

$$
x(x+1)(x+2)(x+3)=y(y+1).
$$
If we perform the multiplication on both sides of the equality and add $l$ to both sides, then on the left side we have $x^{4}+6 x^{3}+11 x^{1}+6 x+1=\left(x^{2}+3 x+1\right)^{2}$, which is a perfect square. Thus, the right side, $\left(y^{2}+y+1\right)$, must also be a perfect square. If $y \geqq 1$, then

$$
y^{2}y^{2}+y+1>(y+1)^{2}
$$

$y^{2}+y+1$ falls between two consecutive perfect squares in both cases, so it cannot be a perfect square.

This leaves the cases $y=0$ and $y=-1$. In both cases, the right side is zero, so the left side must also be zero. The possible values of $x$ are thus $0, -1, -2$, and -3. Therefore, the equation has eight integer $(x, y)$ pairs that satisfy it:

$$
\begin{gathered}
(0,0),(-1,0),(-2,0),(-3,0) \\
(0,-1),(-1,-1),(-2,-1),(-3,-1)
\end{gathered}
$$

These eight pairs are indeed solutions to the equation.