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 $u^2-pv^2 = \pm 1$.)
$x^2+23y^2=2z^2\iff x^2+(5y)^2=2(z^2+y^2)$.
Since the solutions of the equation $X^2+Y^2=2Z^2$ are given by the identity
$$(a^2+2ab-b^2)^2+(a^2-2ab-b^2)^2=2(a^2+b^2)$$ we can try $(y,z)=(a,b)$ taking care of one of $a^2+2ab-b^2$ or $a^2-2ab-b^2$ be equal to $5b$.
Taking for example $(y,z)=(1,4)$ we get the solution $(x,y,z)=(3,1,4)$.
Thus the proposed equation have solutions (which can be parametrized but I stop here).
Best Answer
Continuing from @Batominovski's answer,
the problem asks us to find all the values taking $\pm 1$ in the three following recurrent linear sequences satisfying $a_n = 5a_{n-1} - 7a_{n-2}$,
Looking at the recurrence relation mod $18$, we have $a_n \equiv 5a_{n-1}+11a_{n-2} \equiv 5(5a_{n-2}+11a_{n-3}) + 11a_{n-2} \equiv a_{n-3}$.
Looking at the first three terms of each sequence modulo $18$, we can already discard $2/3$ of all the terms and also remove the possibility of a $-1$ appearing, which reduces the problem to finding the $1$ values in the linear recurrent sequences satisfying $a_n = 20a_{n-1}-343_{n-2}$
Now each sequence is a linear combination of the coefficient sequences of $(2-\omega)^{3n} = (1-18\omega)^n = (10 - 9 \sqrt{-3})^n$, that is, the two sequences
(with integer coefficents in all three cases)
With the binomial theorem, we can expand $(1+9(1-\sqrt{-3}))^n = 1 + 9(1-\sqrt{-3})n + 81(1-\sqrt{-3})^2n(n-1)/2 + \ldots$.
If we place ourselves in $\Bbb Z_3[\sqrt{-3}]$ and we notice that $v_3(9^k/k!) >= 3k/2 \to \infty$, we can reorder the summation and get $1 + [9(1-\sqrt{-3}) + O(81)]n + O(81)n^2 + \ldots$ i.e. each coefficient sequence can be extended as a function $\Bbb Z_3 \mapsto \Bbb Z_3$ which is a power series in $n$ with coefficients in $\Bbb Z_3$
Now, if a power series is of the form $a_0 + a_1 n + a_2n^2$ with $|a_1| > |a_2|,|a_3|,|a_4|,\ldots$, then it is a bijection from $\Bbb Z_3$ to $a_0+a_1\Bbb Z_3$.
The $a_1$ coefficients modulo $81$ in our three sequences are $9-9=0, 9-9 \times (-5) = 54 \neq 0, 9-9 \times (-1) = 18 \neq 0$.
In the last two cases we deduce that the sequence is injective so the initial occurence of $1$ is the only one.
As for the first sequence, we have that both $a_1$ and $a_2$ are of the order of $3^4$. But then its derivative is of the right form to show that it has a single zero at some $n_0 \in \Bbb Z_3$. By setting $m = n-n_0$, we get a new power series of the form $b_0 + b_2m^2 + b_3m^3 + \ldots$ where $|b_2| > |b_3|,|b_4|,\ldots$. This shows that the power series is $2$-to-$1$ (with the exception of $0 \mapsto b_0$), so the two values of $1$ we already know are the only ones.