In my solution to this MSE problem, I noted that $2x^2-1=y^5$ is unlikely to have solutions in integers with $y>1$. Recently, I've tried to find a proof, without success.
Following Thomas Andrews's suggestion (see his solution to the same problem), we start by writing $1-2x^2=(-y)^5$. Working in the ring $\mathbb{Z}[\sqrt{2}]$ we can factorize this as $(1-x\sqrt{2})(1+x\sqrt{2})=\text{ a fifth power }.$ After showing that $1-x\sqrt{2}$ and $1+x\sqrt{2}$ are relatively prime (always working in $\mathbb{Z}[\sqrt{2}]$), the proposed equation would force $1+x\sqrt{2}=u(a+b\sqrt{2})^5$ for some integers $a,b$ and some unit $u$.
The group of units in $\mathbb{Z}[\sqrt{2}]$ is given by $\pm (1+\sqrt{2})^n$ for $n\in \mathbb{Z}$. A minus sign could be absorbed into the fifth power, so we only need to show that $1+x\sqrt{2}=(1+\sqrt{2})^\delta(a+b\sqrt{2})^5$ for some integers $a,b$ and $0\leq \delta\leq 4$ forces $|x|\leq 1$. Here I am following the method described by Adrián Barquero in this MSE question for dealing with an infinite group of units.
For $\delta=0$, it is pretty straightforward but for $\delta=1$, I'd need to show that $1= a^5+20 a^3 b^2+20 a b^4+10 a^4b+40a^2b^3+8b^5$ is not possible, excepting the trivial solution $a=1$ and $b=0$. I haven't made any progress with this, or the other similar expressions we get when $\delta=2,3,4$. I'm stuck here, and this problem seems to me to be as difficult as the original question.
Question: Can the method outlined here be pushed through to give a full solution, or does this problem require higher level approaches?
I'm clearly out of my depth with algebraic number theory, so any suggestions or references that are suitable for amateurs will be greatly appreciated.
Update (Feb. 20) I have followed Ewan's suggestion and tested all the cases $\delta=0,1,2,3,4$ with Sage and there really are no solutions with $y>1$. I will wait a couple of days before rewarding the bounty in case an elementary solution is given. Otherwise, it will go to Ewan.