Solve $x^5 - 1 = y^2$ for integer values of $(x,y)$ . 
Solve $x^5 - 1 = y^2$ for integer values of $(x,y)$ .

What I Tried :- I see that $x^5 = y^2 + 1$ , from here I can  conclude that $x$ has to be positive , because if $x \leq 0$ , then $x^5 \leq 0$ , but $y^2 + 1 > 0$.
Also I thought that maybe $(x^\frac{5}{2} + 1)(x^\frac{5}{2} - 1) = 1$ would do the trick [which would have implied that both the terms are $(1,1)$ or $(-1,-1)$], but that dosen't necessarily mean that $x^\frac{5}{2}$ is an integer .
Then I see that :-
$x^5 - 1 = y^2$
$\rightarrow (x-1)(x^4 + x^3 + x^2 + x + 1) = y^2$
From here the only idea is that $x \neq 2$ , because if it is then it can't be a perfect square (it's a foolish idea though) .
But I tried playing this problem with many ways , but I am stuck at the same place . Can anyone help ?
Note :- It's given that answer is only $(1,0)$ , but how is it coming?

So
$$x^5=y^2+1=(y+i)(y-i).$$
Considering this modulo $4$ gives $y$ even and $x$ odd.
The gcd of $y\pm i$ in the Gaussian integers divides $2$ and $y^2+1$ (which is odd)
so it is $1$. As the Gaussian integers has unique factorisation
and has four units,
both $y\pm i$ are fifth powers, so
$$y+i=(a+bi)^5=(a^5-10a^3b^2+5ab^4)+(5a^4b-10a^2b^3+b^5)i.$$
So $b\mid 1$ and $b=\pm1 $ and
$$5a^4-10a^2+1=\pm1.$$
The only integer solution of this is $a=0$ leading to $y=0$ and $x=1$.
