Is there an elementary proof that $y^2=8x^4+1$ has no integral solution for $x\ge2$? 
How can I prove that $y^2=8x^4+1$ has no integral solution with $x\ge 2$ with elementary methods ?

With elementary I mean using only modular arithmetic, the unique factorization theorem and the theory of quadratic residues modulo a prime $p$.
I tried various approaches , but I did not manage to prove the claim :
First of all, if $x\ge 2$, there is a prime factor $p$ dividing $x$.
$y^2-1=(y-1)(y+1)=8x^4$
So, if $y-1$ or $y+1$ is divisible by an odd prime, we have $p^4|y-1$ respective $p^4|y+1$. Does this lead to anywhere ?
$y^2-9=8x^4-8$ , which implies $(y-3)(y+3)=8(x-1)(x+1)(x^2+1)$. Can I make use of the fact that every odd prime $p$ dividing $x^2+1$ has the form $4k+1$ ?
Since the equation is closely related to the triangular numbers (The question whether a triangular number can be a fourth power leads to the given equation), I also studied the convergents of $\sqrt{2}$ (which satisfy the pell-equation $x^2-2y^2=1$). 
The sequence $B_n$ of the positive integers, whose squares are triangular satisfy the recurrence relation $B_1=1$ , $B_2=6$ , $B_n=6B_{n-1}-B_{n-2}$. So, showing that no $B_n$ except $1$ is a perfect square would also finish the proof.
Any ideas ?
 A: The positive integer $y$ must be odd, and letting $z=(y-1)/2$ we get $z(z+1)=2x^4.$
The two consecutive integers $z$ and $z+1$ have no common prime factors, so
one is a fourth power and the other twice a fourth power. 
Case 1: If $z=\ell^4$ and $z+1=2m^4$, then $\ell^4+1=2m^4$.
  From my answer here,
 we see that $\ell=1$ and so $y=3$ and $x=1$.
(This part uses the fact that $z^2=x^4-y^4$ has no solutions in non-zero integers.
This is Exercise 1.6 in Edwards's book on Fermat's Last Theorem.
The proof uses the representation of Pythagorean triples and infinite
descent.) 
Case 2: If $z=2m^4$ and $z+1=\ell^4$, then $\ell^4-1=2m^4$. 
Since $(\ell^2-1)(\ell^2+1)=2m^4$ and $\gcd(\ell^2-1,\ell^2+1)\leq 2$, one of the factors $\ell^2-1$ or $\ell^2+1$ is a fourth power, in particular a square.
The only two consecutive squares are $0$ and $1$, so we must have $\ell=1$.
This implies $z=0$ and so $y=1$ and $x=0$. 
A: Here is another approach. We can rewrite original equation as 
$y^2 - 1 = 2^kx^4$  where $x$ is odd. Equation (1)
$(y-1)(y+1) = 2^k m^4 n^4$ where $(m,n) = 1$ and both $m$ and $n$ are odd. 
$gcd(y-1, y+1) = 2$
STEP 1: Without loss of generality, we can write
$y - 1 = 2^{k-1} m^4$ and $y + 1 = 2n^4$. The difference gives us,
$n^4 - 2^{k-2} m^4 = 1$ 
We have an equation that looks like Equation(1) with $n^2 < y, m < x$ and so this descent will lead us to an equation of the form $r^4 - 2s^4 = 1$ Equation (2)
or $r^4 - s^4 = 1$ Equation (3)
with gcd(r,s) = 1 and $r, s$ being odd
The LHS and RHS of Equation (2) (and Equation (3)) are not congruent mod 4. 
Hence Equations (2) and (3) have no solution besides the trivial solution $r = 1, s = 0$
Hence Equation (1) has no solution in integers besides $x = \pm1$
ps: If in STEP 1, we interchanged the values of $y-1$ and $y+1$, we will get
$2^{k-2}m^4 - n^4 = 1 \implies 2^{k-2}m^4 = (n^2)^2 + 1^2$
Since $k$ is even, this is of the form $p^2 = n^4 + 1^4$ which do not have solutions in integer $p, n$ per Fermat. 
