The Diophantine equation $x^5-2y^2=1$ I'm trying to solve the Diophantine equation $x^5-2y^2=1$.
Here's my progress so far. We can write the Diophantine equation as
$$\frac{x-1}{2}\cdot(x^4+x^3+x^2+x+1)=y^2.$$
If $x\not\equiv1\pmod{5}$, then $\gcd(\frac{x-1}{2},x^4+x^3+x^2+x+1)=1$, so both $\frac{x-1}{2}$ and $x^4+x^3+x^2+x+1$ must be perfect squares (note: $x^4+x^3+x^2+x+1>0$).
In particular, $4(x^4+x^3+x^2+x+1)$ is a perfect square.
Comparison with $(2x^2+x)^2$ and $(2x^2+x+1)^2$ forces $-1\leq x\leq3$.
This results in the solutions $(3,\pm11)$.
If $x\equiv1\pmod{5}$, then we can write the Diophantine equation as
$$\frac{x-1}{10}\cdot\frac{x^4+x^3+x^2+x+1}{5}=\left(\frac{y}{5}\right)^2,$$
where $\gcd(\frac{x-1}{10},\frac{x^4+x^3+x^2+x+1}{5})=1$, so both $\frac{x-1}{10}$ and $\frac{x^4+x^3+x^2+x+1}{5}$ must be perfect squares.
Thus,
$$x=10a^2+1,$$
$$x^4+x^3+x^2+x+1=5b^2.$$
Unfortunately, this is where I get stuck.
I can substitute the first equation into the second, giving
$$10000a^8+5000a^6+1000a^4+100a^2+5=5b^2,$$
$$2000a^8+1000a^6+200a^4+20a^2+1=b^2,$$
$$2000a^8+1000a^6+200a^4+20a^2=(b-1)(b+1),$$
$$5a^2(100a^6+50a^4+10a^2+1)=\frac{b-1}{2}\cdot\frac{b+1}{2},$$
but this doesn't seem to be making progress, even with modular arithmetic considerations.

Here is an "elementary" proof. The given diophantine equation $x^5 = 1+2y^2$ admits the obvious solution $x=1, y=0$. Exclude this trivial solution and consider $a=x^5$ as an integral parameter which one wants to represent as the value of the quadratic form $t^2+2y^2$, with unknown integers $(t,y)$. Geometrically, the problem is equivalent to find the points of the sublattice $\mathbf Z^2$ of $\mathbf R^2$ which belong to the ellipse with equation $a= t^2+2y^2$. Since the lattice is discrete and the ellipse is compact, the set $S$ of wanted points is finite. If $S$ is not empty and $t=1$, symmetry w.r.t. the $t$-axis imposes that card $S=2$. If one follows this elementary approach, the only reason for the hypothesis $a=x^5$ seems to be the quick growth of the 5-th power, which allows to determine $S$ without too many trials.
NB : As usual, "elementary" methods often mask the power - and slickness - of more  elaborate methods. The general process at work here is, as suggested by the answer given by @leoli1, the representation of a positive integer by a binary quadratic positive definite form.
