Solve the equation $(x+y)^2 + 3x + y + 1=z^2$ over positive integers. 
Solve the equation $(x+y)^2 + 3x + y + 1=z^2$ where $x, y, z \in
 \mathbb{N}$

I've found some solutions, like $(0, 0, 1), (1, 1, 3)$ and, more general, $x=k,y=k,z=2k+1$. No idea how to prove or disprove there is no other solutions.

If we set $x+y=t$, we are just looking for the solutions of 
$$ (2t+1)^2+(8x+3) = 4z^2 \tag{1}$$
or:
$$ (8x+3) = (2z+2t+1)(2z-2t-1).\tag{2} $$
If $d$ is a divisor of $8x+3$ that is $\leq \sqrt{8x+3}$, $d$ is odd and
$$ 2z+2t+1=\frac{8x+3}{d},\qquad 2z-2t-1 = d \tag{3} $$
leads to the solution:
$$ z = \frac{1}{4}\left(d+\frac{8x+3}{d}\right),\qquad 2t+1=\frac{1}{2}\left(\frac{8x+3}{d}-d\right).\tag{4} $$
I leave to you to fill the missing details, like, for instance, to find the conditions that ensure the last $z$ and $t$ actually being positive integers with $t\geq x$.
