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?
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.
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 $(x,y,z)=(7,1,55)$