Prove that there are no integer solutions to: $x^4+6x^2+1=8y^4$ 
Prove that there are no integer solutions to:
$$x^4+6x^2+1=8y^4$$
where $x>1$.

My attempts:
Let $x^2=u, y^2=v$
$$u^2+6u+(1-8v^2)=0$$
$$\Delta=36-4(1-8v^2)=w_0^2$$
$$32v^2+32=(4w_1)^2$$
$$2v^2+2=w_1^2=(2w_2)^2$$
$$v^2+1=2w_2^2$$
$$v=w_2=1\implies x=y=1$$
I don't know how can I proceed.
 A: First, observe that $8y^4$ is even, so $x^4+6x^2+1$ is even, so $x^4+6x^2$ is odd. Therefore, $x$ is odd. Rewrite $x$ as $2n+1$, where n is an integer and $n\geq0$. So the original equation can be rewritten as $(2n+1)^4+6(2n+1)^2+1=8y^4$. By simplifying it we see that $2n^4+4n^3+6n^2+4n+1=y^4$, which can be rewritten as $(n+1)^4+n^4=y^4$, where $n$ and $y$ are integers. It's obvious that when $n=0$ and $y=1$ the equation holds, but since $x>0$ this cannot be the case. So $n\geq1$. But by Fermat's Last Theorem, such $n$ and $y$ does not exist.
A: Using $v = y^2$ in your discriminant result, we have
$$v^2 + 1 = 2w_2^2 \; \; \to \; \; y^4 + 1 = 2w_2^2 \tag{1}\label{eq1A}$$
As explained in the accepted answer to Solve in $\mathbb{Z}$ the equation $x^4 + 1 = 2y^2$. (with the comment there linking to the answer re: the solutions of the transformed $X^4 - Y^4 = Z^2$), the only solution to \eqref{eq1A} is with $y = \pm 1$, so this means $x = \pm 1$. As such, there are no integer solutions with $x \gt 1$.
A: $x^4 + 6x^2 + 1 = (x^2 + 3)^2 + 1 - 9 = 8y^4 \implies (x^2 + 3)^2 = 8 + 8y^4 = (2y^2 + 2)^2 + (2y^2 - 2)^2$
So the three numbers $a = 2y^2 + 2, b = 2y^2 - 2, c = x^2 + 3$  must be a pythagorean triple and for this to be true, $a,b,c$ should be in the form $m^2 - n^2, 4mn, m^2 + n^2$ which is not the case.
