Does the equation $x^2+23y^2=2z^2$ have integer solutions?
$x^2+23y^2=2z^2\iff x^2+(5y)^2=2(z^2+y^2)$.
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).
Conclusion: it is likely that the parametrization $$ \color{blue}{ \left(7u^2 + 10uv -3 v^2 \right)^2 + 23 \left(u^2 -2uv - v^2 \right)^2 = 8 \left( 3u^2 + uv + 2 v^2 \right)^2} $$ gives all solutions to $x^2 + 23 y^2 = 8 z^2$ with $\gcd(x,y,z) = 1.$ There is a theorem that all primitive solutions can be found with a small finite number of such parametrizations. This version is better than the one at the end of my discussion, $z$ is discriminant $-23$ rather than $-92,$ that is why there is no longer a problem about $z \equiv 2 \pmod 4.$ Live and learn.
If you wish to see how this does by computer (I advise this), with $z = 3u^2 + uv + 2 v^2 \leq M$ for some upper bound $M,$ we can demand $$ |u| \leq \sqrt {\frac{8M}{23}}, $$ $$ |v| \leq \sqrt {\frac{12M}{23}}. $$
Here is a raw search for primitive solutions with $z \leq 100.$
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x y z
3 1 2
7 1 3
25 1 9
5 7 12
15 7 13
45 1 16
59 7 24
9 17 29
81 7 31
61 17 36
1 23 39
111 7 41
105 17 47
147 1 52
149 7 54
35 31 54
135 31 71
53 41 72
63 41 73
205 17 78
163 31 78
41 47 81
73 49 87
263 1 93
259 17 96
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Next is by parametrization, where it was necessary to use absolute values to get a nice ordering. In order to reduce repetition I took only $u \geq 0.$ I notice how not all $\pm$ signs occur in $x,y$ here. To get all types, one would need four slightly different parametrizations, $(x,y)$ as written, then $(x,-y),$ $(-x,y),$ $(-x,-y).$
z |x| |y| z x y z u v
2 3 1 2 -3 -1 2 0 1
2 3 1 2 -3 -1 2 0 -1
3 7 1 3 7 1 3 1 0
9 25 1 9 -25 1 9 1 -2
12 5 7 12 5 7 12 2 -1
13 15 7 13 15 -7 13 1 2
16 45 1 16 45 -1 16 2 1
24 59 7 24 -59 7 24 2 -3
29 9 17 29 -9 17 29 3 -2
31 81 7 31 -81 -7 31 1 -4
36 61 17 36 61 -17 36 2 3
39 1 23 39 -1 -23 39 1 4
41 111 7 41 111 -7 41 3 2
47 105 17 47 -105 17 47 3 -4
52 147 1 52 -147 -1 52 2 -5
54 149 7 54 149 7 54 4 1
54 35 31 54 -35 31 54 4 -3
71 135 31 71 135 -31 71 3 4
72 53 41 72 53 -41 72 2 5
73 63 41 73 63 41 73 5 -2
78 163 31 78 -163 31 78 4 -5
78 205 17 78 205 -17 78 4 3
81 41 47 81 -41 -47 81 1 6
87 73 49 87 -73 49 87 5 -4
93 263 1 93 263 1 93 5 2
96 259 17 96 -259 -17 96 2 -7
z |x| |y| z x y z u v
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There is another way to talk about the behavior of the norm map. The binary quadratic forms of discriminant $-92$ are in three classes, reduced are $x^2 + 23 y^2,$ $3x^2 + 2xy + 8y^2,$ $3x^2 - 2xy + 8y^2.$
That is, under Gauss composition, the groups is cyclic order three and $3x^2 + 2xy + 8y^2$ is a generator. It primitively represents $8.$ It also primitively represents infinitely many squares $z^2,$ and composition tells us that $8z^2$ is primitively represented by $x^2 + 23 y^2.$
The part about squares is best written in Dirichlet's version of composition, which was all we had until Bhargava gave other interpretations of Gauss. The form $3x^2 + 2xy + 8y^2$ is integrally equivalent ($SL_2 \mathbb Z$) to $3x^2 + 4 xy + 9 y^2.$
Dirichlet's methods tell us (page 49 in Cox, although the first edition has a typo, pasting corrected selection from second edition ) that $$ \left( 3 u^2 + 4 uv + 9 v^2 \right)^2 = 9 U^2 + 4 UV + 3 V^2, $$ where $$ U = u^2 - 3 v^2, \; \; \; V = 6uv + 4 v^2. $$
Another arrangement of the symbols $a,a',B,C$ tells us $$ \left( 9U^2 + 4 U V + 3 V^2 \right) \left(3 s^2 + 4 st + 9 t^2 \right) = 27 S^2 + 4 ST + T^2, $$ where $$ S = Us-Vt, \; \; \; T = 9Ut + 3Vs +4Vt. $$ Taking lower case $s=1,t=-1$ gives $$ 8 \left(3 U^2 + 4 UV + 9 V^2 \right) = 27 S^2 + 4 ST + T^2, $$ where $$ S = U +V, \; \; \; T = -9U + 3V - 4V = -9U -V. $$ Finally $$ (T+2S)^2 + 23 S^2 = 27 S^2 + 4 ST + T^2. $$
Put it together, we can, for example with $p$ prime, solve $x^2 + 23 y^2 = 2 p^2$ whenever $p = 3 u^2 + 2 uv + 8 v^2.$ This is possible for $p = 3$ and then for all $(23|p) = 1$ such that $ w^3 - w + 1$ is irreducible $\pmod p.$ This last fact is not in Cox, it is class field theory from a 1991 article by Hudson and Williams.
Maybe I should summarize: we get an explicit formula for $x,y$ in $$ x^2 + 23 y^2 = 8 \left( 3 u^2 + 4 uv + 9 v^2 \right)^2 $$ given $u,v$ arbitrary integers. That is, $$ \color{blue}{ \left(-7u^2 + 6uv + 25 v^2 \right)^2 + 23 \left(u^2 + 6uv + v^2 \right)^2 = 8 \left( 3u^2 + 4 uv + 9 v^2 \right)^2}. $$ Now that I've corrected some errors of mine (I was out of practice) the final formula can be checked easily enough.
