Integer solutions to $x^2=2y^4+1$. Find all integer solutions to $x^2=2y^4+1$.

What I tried 
The only solutions I got are $(\pm 1 ,0)$, I rewrote the question as : is $a_{n}$ a perfect square for $n>0$ were 
$$a_0=0,\quad a_1=2, \quad a_{n+2}=6a_{n+1}-a_n.$$ 
I tried taking $\pmod{4}$ and $\pmod{12}$ but that lead me nowhere.

There are no solutions other than those trivial ones. First we note that $x$ must be odd. Let $x = 2k+1$. This gives $2k^2+2k = y^4$. Thus $y = 2j$ and we have $k(k+1) = 8j^4$. Note that $\gcd(k,k+1) = 1$ so we have two cases.
Case 1: $k+1 = 8a^4, k = b^4$.
This case is impossible since this implies $b^4 \equiv -1 \pmod 4$ but $-1$ is a not a quadratic residue $\pmod 4$.
Case 2: $k+1 = b^4, k = 8a^4$.
This case is equivalent to solving the diophantine equation $8a^4+1 = b^4$. An elementary solution is given here. In the referenced paper, it is prove that the only non trivial solution to this equation is $(a,b) = (1,3)$ which gives $k = 8, k+1 = 9$ but $9 = b^4$ is impossible.
