Integer solutions of $ x^4+34x^2 y^2+y^4=z^2$ Now I'm trying to solve one problem, and already twice an expression of the form $a x^4+b x^2 y^2+a y^4 \,\,(x > y > 1)$ appears under the square root.
First was: $9 x^4 - 14 x^2 y^2 + 9 y^4$
Second: $x^4 + 34 x^2 y^2 + y^4$
I suppose that there will be other equations of this type.
Question: Are there integer solutions or a parametrization of equations$ a x^4+b x^2 y^2+a y^4=z^2$ in general, or of the variants proposed above in particular? Or maybe can the absence of solutions be proven?
Upd: No solutions
Second case: according to https://www.cambridge.org/core/services/aop-cambridge-core/content/view/0B14DEDE386776126F3B5A36CA325ECF/S0017089500007862a.pdf/x4-dx2y2-y2-z2-some-cases-with-only-trivial-solutions-and-a-solution-euler-missed.pdf (Thanks Will Jagy for link)
First case: $9 x^4 - 14 x^2 y^2 + 9 y^4$ can be written as  $(3(x^2-y^2))^2 +(2xy)^2=z^2$  Pythagorean triple. So $2xy=2ab, 3(x^2-y^2)=a^2-b^2$
From this: $3y^4+(a^2-b^2)y^2-3a^2b^2=0$ can be obtain.
Discriminant $a^4+34a^2b^2+b^4$ (Second equation of my question. Strange coincidence). So there is no rational roots.

At first, it looks like the equation can be rewritten as a Pythagorean triple
$\space A^2+B^2=C^2\quad$ here shown by Euclid's formula:
$\quad A=(m^2-k^2) \quad B=2mk\quad C=(m^2+k)^2\space $ and a triple may be found given any legitimate value of $A,\space B, \space \text{or}\space C.$
$\text{Here we have}\quad(x^2-y^2)^2 + (6xy)^2 =z^2\space$ and $\space B= 6xy\ne 2mk\space$  so it is not a Pythagorean triple because it cannot quite be a right triangle. The best approach may be experimentation and observation.
\begin{align*}
x=0,\space y=0 &\implies z\in\mathbb{Z}^2
\\
x=y,\space\space \space  x,y>0 &\implies z=6n^2\space ,n\in\mathbb{N}\\
\not\exists \space x>y>1
\\
x=0,\space y=4 &\implies z=\pm8
\\
x=4,\space y=0 &\implies z=\pm8
\end{align*}
