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.

Write this as 
$$3x + y + 1 = z^2 - (x+y)^2 = (z-x-y)(z+x+y)$$
For any integers $a,b$, we might look for a solution with
$$ \eqalign{a &= z - x - y\cr
            b &= z + x + y\cr
           ab &= 3x + y + 1\cr} $$
Solving this system for $x,y,z$:
$$ \eqalign{x &= \frac{ab}{2} + \frac{a}{4} - \frac{b}{4} - \frac{1}{2}\cr
            y &= \frac{-ab}{2} - \frac{3a}{4} + \frac{3b}{4} + \frac{1}{2}\cr
            z &= \frac{b}{2} + \frac{a}{2}\cr}$$
In order for $x, y, z \ge 0$, we need $b > 0$ with 
$$ \frac{b+2}{2b+1} \le a  \le \frac{3b+2}{2b+3} $$
The only integer that fits is $a = 1$.  Then for $x$ to be an integer we need $b \equiv 1 \mod 4$.  The result is that all the positive integer solutions are the ones you found.
