Update: It only remains to handle Equation $x^2-y^5=2$ (see edit below).
Question 1. What are the integer solutions to the equations $x^2-y^5=\pm2$?
I know that these diophantine equations have only finitely many solutions, but I don't understand the subject well enough to give a good argument. I suspect this follows from Siegel's Theorem and involves the notion of genus, but I'm not sure, so let me ask the question:
Question 2. What would be a nice way to see (possibly using known results) this finiteness result?
(I'm not planning to accept an answer solving only Question 2.)
Equation $x^2-y^5=2$ has the solutions $(\pm1,-1)$. I don't know any other solution, and I don't know any solution to $x^2-y^5=-2$.
To find the non trivial solutions (if any) we can clearly assume $x,y\ge3$.
The only thing I've been able to show is that any non trivial solution satisfies $y>593825$.
To verify this I use this answer of Max Alekseyev. Indeed, $x^2$ and $y^5$ are two powerful numbers which differ by $2$. The first $13$ such pairs of powerful numbers are given in this OEIS entry, and it is easy that check that none of them works. The largest number occurring in these $13$ pairs is
$$
73840550964522899559001927225
$$
and the integer part of its fifth root is $593825$.
I think it would be easy for somebody who (unlike me) knows how to use the appropriate softwares to improve this bound, and I hope some users will do that if answering Question 1 turns out to be hard.
EDIT. Thank you to Dietrich Burde who pointed out in a comment that Ljunggren proved that Equation $x^2-y^5=-2$ has no integer solution. Dietrich linked to the article
Abu Muriefah, Fadwa S., and Yann Bugeaud. The diophantine equation $x^2+c=y^n$: a brief overview. Revista Colombiana de Matemáticas 40.1 (2006): 31-37.
The article mentions also subsequent papers of Nagell and Sury which also prove the result (Sury's approach being more elementary).
As indicated at the beginning of the post, it only remains to handle Equation $x^2-y^5=2$.
UPDATE. I just noticed that the article
Samir Siksek. "The modular approach to Diophantine equations." Number Theory: Volume II: Analytic and Modern Tools (2007): 495-527, PDF,
contains an answer to the question. More precisely, it shows that the only solutions to $x^2-2=y^p$, with $p$ an odd prime, are $(x,y,p)=(\pm1,-1,p)$ if $p$ is not in the range
$$
41\le p\le 1231.
$$
There are a little more details at the end of this answer.
Best Answer
Let $x$ and $y$ be integers such that $$x^2-y^5=2.$$ Then in the ring $\mathbb{Z}[\sqrt{2}]$, which is a UFD with unit group $\langle-1,1+\sqrt{2}\rangle$, we have the identity $$y^5=x^2-2=(x+\sqrt{2})(x-\sqrt{2}).$$ The two factors on the right hand side are coprime because $x$ is odd, and so both factors are fifth powers, up to units. That is to say, we have $$x+\sqrt{2}=\pm(1+\sqrt{2})^k(a+b\sqrt{2})^5,$$ for some $k\in\{0,1,2,3,4\}$. Note that $-1$ is a fifth power, so we may omit the $\pm$-sign. We have $$(a+b\sqrt{2})^5=a(a^4+20a^2b^2+20b^4)+b(5a^4+20a^2b^2+4b^4)\sqrt{2},$$ so for $k=0$ we immediately see that $b=\pm1$ and so $$5a^4+20a^2+4=b=\pm1,$$ reducing mod $5$ shows that $b=-1$, but this does not yield any integer solutions for $a$.
By a similar calculation, for $k=1$ we find that $$1 =a^5+5a^4b+20a^3b^2+20a^2b^3+20ab^4+4b^5.$$ This is a Thue equation, for which effective methods to solve them exist. PARI/GP tells me that the only solution is the trivial solution $(a,b)=(1,0)$, corresponding to $(x,y)=(1,-1)$.
The exact same can be done for $k=2,3,4$. Each time you will find a quintic Thue equation, and a solver will list all integral solutions. There are guaranteed to be only finitely many in each case. For $k=2,3,4$ I find \begin{eqnarray} 1&=&2a^5+15a^4b+40a^3b^2+60a^2b^3+40ab^4+12b^5,\\ 1&=&5a^5+35a^4b+100a^3b^2+140a^2b^3+100ab^4+28b^5,\\ 1&=&12a^5+85a^4b+240a^3b^2+340a^2b^3+240ab^4+68b^5. \end{eqnarray} Only the latter has an integral solution, which is $(a,b)=(-1,1)$. This corresponds to the trivial solution $(x,y)=(-1,-1)$.
This method is guaranteed to give all finitely many solutions within a reasonable amount of time though.