Find all triplets of strictly positive integers $(x, y, z)$ such that

$$
(x+y)^{2}+3 x+y+1=z^{2} .
$$
We can easily see that

$$
(x+y)^{2}<(x+y)^{2}+3 x+y+1<(x+y+2)^{2}
$$

Therefore, the only solution is $z=x+y+1$. We have

$$
(x+y)^{2}+3 x+y+1=(x+y+1)^{2}
$$

and we find $x=y$. The solutions are therefore the triplets of the form $(x, x, 2 x+1)$.