Considering this as a quadratic in $y$, it is sufficient (and necessary) that
$n^4 - 4n^2(x^2-z^2)$ is a perfect square.
i.e.
$n^2 - 4(x^2 - z^2)$ is a perfect square, say $q^2$.
Rewriting gives us
$(n-q)(n+q) = 4 (x^2 - z^2)$
If $n=2k$ is even, then $q$ needs to be even too (say $2m$) and
$(k-m)(k+m) = (x-z)(x+z)$
If $n=2k+1$ is odd, then $q$ needs to be odd too (say $2m+1$) and
$(k-m)(k+m+1) = (x-z)(x+z)$
Thus you can pick any $m$, and try to factor the left hand side above (choosing the right one depending on the parity of $n$) into two terms of the same parity.
This is an old post but may be useful to those who come across it. The problem can be reduced to an identity. Given,
$$ax^2+bxy+cy^2+dx+ey+f=0\tag{1}$$
for integer constants $a,b,c,d,e,f$. First, solve it as an eqn in $y$,
$$y = \frac{-e-bx \pm \sqrt{ px^2+qx+r}}{2c}\tag{2}$$
where,
$$p,\;q,\;r = (b^2-4ac),\; -2(2cd-be),\;(e^2-4cf)\tag{3}$$
Thus, if the discriminant of $(2)$ is a square, or you have an initial rational solution to,
$$px_0^2+qx_0+r = t^2\tag{4}$$
then it implies $(2)$ is rational.
So here is the relevant identity. Let $p,q,r$ be defined as above. Then,
$$\begin{aligned}
&ax^2+bxy+cy^2+dx+ey+f=\frac{(px_0^2+qx_0+r)-t^2}{-4c}=0\\
\text{where},\qquad\\
&y = \frac{-e-bx \pm \big(u/v(x-x_0)+t\big)}{2c}\\
&x = x_0+\frac{-2tuv+(2px_0+q)v^2}{u^2-pv^2}
\end{aligned}\tag{5}$$
for arbitrary $u,v$. So if you have initial rational solution $x_0$ to $(4)$, then the identity $(5)$ shows you can generate an infinite more.
(P.S. Furthermore, if $c=1$, and non-square $p=b^2-4ac>0$, then you can find integer $x,y$ by solving the Pell equation $u^2-pv^2 = \pm 1$.)
Best Answer
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.