Find all positive integer pairs $(x, y)$ such that $x^{2}=y^{2}+7 y+6$.
Let $(x, y)$ be a potential solution. For $y>3$, we have $(y+3)^{2}=y^{2}+6 y+9<y^{2}+7 y+6=$ $x^{2}<y^{2}+8 y+16=(y+4)^{2}$, which is absurd. Therefore, $y \in \{0, 1, 2, 3\}$. By testing the three cases, only $y=3$ gives a perfect square $(x=6)$.

The only solution is therefore $(6,3)$.