Upper bound on the number of integer solutions to $y^p=x^2+2$, where $p$ is prime

Question

I need to find an upper bound on the number of solutions of the Diophantine equation $y^p=x^2+2$, where $p$ is prime.
I have previously considered the equation $y^3=x^2+2$ and proved its solutions are $(5,3)$ and $(-5,3)$ by factorizing it as $y^3=(x-\sqrt-2)(x+\sqrt-2)$ and showing that $(x-\sqrt-2)$ and $(x+\sqrt-2)$ are relatively prime by properties of the UFD $\mathbb{Z}[\sqrt-2]$ and then showing that $x+\sqrt-2$ is a cube.
In the last stept, I have used the binomial theorem to find the solutions:
$x+\sqrt-2=(a+b\sqrt-2)^3$, where $a,b \in \mathbb{Z}$.
I think I need to do something similar for $y^p=x^2+2$. Obviously, I can show that $x+\sqrt-2$ is a $p$th power and then use the binomial theorem to get:
$x+\sqrt-2=(a+b\sqrt-2)^p=\sum_{k=0}^{p} \binom{p}{k} a^{p-k} (b\sqrt-2)^k$.
How do I obtain an upper bound on the number of solutions of the equation from here?

Answer 1

This has been discussed in the article here on the Diophantine equation $$ y^n=x^2+c $$ for a integers $n$ and $c$, so in particualr for $c=2$ and $n=p$ prime. Ljunggren [20] generalised Fermat’s result and proved that for $c = 2$ the equation has no solution other than $x = 5$.

Other references on this site: the number of integer solutions to $y^p = x^2 +4$