Three variable, second degree diophantine equation I am trying to solve this diophantine equation:
    $x^2 + yx + y^2 = z^2$
In other words, I am trying to find integers $x$ and $y$ such that $x^2 + yx + y^2$ is a perfect square.
So far, the only methods to solve quadratic diophantine equations I am familiar with are Pythagorean triples and Pell equations. The $yx$ term has a coefficient of one, so I can't complete the square to reduce it to a Pell equation somehow, and I am wondering if there are other methods to solve this kind of equation.
Some insight would be highly appreciated.

Let $z$ be a fixed integer. We have $x^2+y^2+xy=z^2$ is equivalent to 
$(x+y)^2-xy=z^2$ and $(x-y)^2+3xy=z^2.$ Therefore, $(x+y)^2=z^2+xy\geq 0$ and $(x-y)^2=z^2-3xy\geq 0.$ Then, we have $z^2+xy$ and $z^2-3xy$ are squares.
So, $|x+y|=\sqrt{z^2+xy}$ and $|x-y|=\sqrt{z^2-3xy}.$ This implies that 
$x+y=\sqrt{z^2+xy}$ or $x+y=-\sqrt{z^2+xy}$  and $x-y=\sqrt{z^2-3xy}$ or $x-y=-\sqrt{z^2-3xy}.$ You use after that the basic tools of calculus to find $x$ and $y$ separatly.
