Find the $(x, y)$ natural integers such that $x^{2}=y^{2}+7 y+6$
We find the order of magnitude of x relative to y:

$(y+2)^{2}=y^{2}+4 y+4$ Therefore $x>y+2$

$(y+4)^{2}=y^{2}+8 y+16$ Therefore $x<y+4$

We deduce that $x=y+3$. It remains to solve $y^{2}+6 y+9=y^{2}+7 y+6$. Thus $y=3$ and then $x=6$. The only solution is therefore $(6,3)$