Solving a homogeneous quadratic equation in three variables over the integers. Is there any procedure for determining if an infinite amount of solutions exist for an equation of the type $x^2 = ay^2 + byz + cz^2$ for arbitrary integer constants $a, b, c$ and variables $x, y, z \in \mathbb{Z_+}$? If not, does knowing at least one non-trivial solution of the equation help determine if an infinite amount of solutions exist?
Example (if it helps): let $x^2 = 202 y^2+14yz+9z^2$. Here one solution is $y=z=1$ and $x=15$, do there exist infinitely many other solutions?
 A: TLDR; If there exists one nonzero solution, then there exist infinitely many solutions, paramatrized by $\Bbb{Z}^2$. Popular candidates to test are triplets $(0,y,z)$ where $y\mid c$ and $z\mid a$, though such solutions need not exist.

Every integral solution to your equation with $x\neq0$ yields a rational solution to
\begin{equation}aY^2+bYZ+cZ^2=1,\end{equation}
by setting $Y:=\tfrac  yx$ and $Z:=\tfrac zx$. Conversely, every rational solution to this equation yields an integral solution to your equation by multiplying out the denominators. This also shows that multiplying an integral solution through by an integer yields another integral solution.
Given a nonzero rational solution $(Y_0,Z_0)$ to your equation, also
$$(aY_0+bZ_0,-aZ_0),$$
is a rational solution, and moreover for every $k\in\Bbb{Z}$ also
$$\left((aY_0+bZ_0)k^2+2cZ_0k-cY_0,-aZ_0k^2+2cY_0k+bY_0+cZ_0\right),$$
is a rational solution. These are in fact all rational solutions. This then in turn yields all integral solutions, except those with $x=0$. For these we have
$$Y=\frac{-b\pm\sqrt{b^2-4ac}}{2a}Z,$$
so such solutions exist if and only if $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ are rational, i.e. if and only if $ax^2+bx+c$ has a rational root, which is easy to test.
A: Given,
$$ay^2 + by z + c z^2 = x^2\tag1$$
for the particular case when $z = 1$, and $a$ is not a square, then YES, given an initial positive integer solution, one can find an infinite more positive ones by solving a Pell equation. For your example,
$$202y^2 + 14y z + 9 z^2 = x^2\tag2$$
the solution is,
$$x = 15 + 418 p q + 6060 q^2$$
$$y = 1 + 30 p q + 418 q^2$$
$$z = 1$$
and $p,q$ solve,
$$p^2-202q^2 = 1$$
This is just a special case of a more general solution.
A: The second option is if you know one any decision.
We need to write generally speaking the more General equation:
$$aX^2+bXY+cY^2=jZ^2$$
If we know what any one solution: $(x,y,z)$  - then you can write a formula for the solutions of this equation.
$$X=jxt^2-cxk^2+2(cyk-jzt)s+(by+ax)s^2$$
$$Y=jyt^2-2jztk+(cy+bx)k^2+2axks-ays^2$$
$$Z=jzt^2-(bx+2cy)kt+czk^2+(bzk-(2ax+by)t)s+azs^2$$
$k,t,s$ - any integer asked us.
A: I think one should first algebraically manipulate it into a form involving a linear equation in whole squares only:
$$
\begin{align}
X ^2 &= a \cdot Y^2 + b \cdot YZ + c \cdot Z^2 \\
\to \\
\vdots \\ 
\to \\
a \cdot \Big[2X\Big]^2+(b^2-4ac)\cdot \Big[Z\Big]^2 &=\Big[2a \cdot Y+b \cdot Z\Big]^2 \\
\end{align}
$$
Then use the time-honored methods of sterographic projection.  This is an example.
For the equation $x^2 = 202 y^2+14yz+9z^2$, $a=202$, $b=14$, $c=9$, gives us in the whole-square equation:
$$808x^2-7076z^2=(404y+14z)^2$$
Reduces to
$$1769z^2+(202y+7z)^2=202x^2$$
Though it looks daunting, you can find all solutions with the one already found by the OP
A: Equation Shown below has parametrization for $(a,b,c,d)=(6,-7,1,54)$
$dx^2=ay^2 + by z + c z^2$
$x=27k^2-20k-7$
$y=135k^2-54k-1$
$z=81k^2+54k-55$
For $k=0$ , we get $(x,y,z)=(7,1,55)$
