Solve $x^2+(8y)^2=p^2(4p^2y^2+1)$

Question

I am trying to find solutions for $x^2+(8y)^2=p^2(4p^2y^2+1)$ for integer $x,y$ where $p$ is a prime $\equiv 1 \mod 4$ that does not divide $x,y$.

I think there are no solutions but I could not prove this. Obviously $x$ is odd, and $4p^2y^2+1$ is a product of primes $\equiv 1 \mod 4$, but I am unable to progress.

Note that the equation can be rearranged to the Pell-like equation $$x^2-(p^4-16)(2y)^2=p^2$$

Answer 1

Try $p=5$, $x=691$, $y=14$. https://www.alpertron.com.ar/QUAD.HTM is a good resource.