<s>[INST] 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.
 [/INST] $x^4 + 6x^2 + 1 = (x^2 + 3)^2 + 1 - 9 = 8y^4 \implies (x^2 + 3)^2 = 8 + 8y^4 = (2y^2 + 2)^2 + (2y^2 - 2)^2$
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.
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