Is there an elementary proof that $y^2=8x^4+1$ has no integral solution for $x\ge2$?

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 $x=0$.