For a prime $p$ and $a,b\in\mathbb{Q}_p^\times$, the Hilbert symbol $(a,b)_p$ is defined to be 1 if $ax^2+by^2=1$ has a solution over $\mathbb{Q}_p$, and $-1$ otherwise.

This is equivalent to the existence of a primitive solution to $z^2-ax^2-by^2=0$ over $\mathbb{Z}_p$.

The value of $(a,b)_p$ depends only on the images of $a$ and $b$ in $\mathbb{Q}_p^\times/\mathbb{Q}_p^{\times 2}$.

For $p=2$ the set $S=\{\pm 1, \pm 2, \pm 3, \pm 6\}$ is a complete set of representatives for $\mathbb{Q}_p^\times/\mathbb{Q}_p^{\times 2}$.

For $u\in\mathbb{Z}_2^\times$, define $\epsilon(u)=(u-1)/2$ and $\omega(u)=(u^2-1)/8$.  Let $a=2^\alpha u$ and $b=2^\beta v$ with $u,v\in\mathbb{Z}_2^\times$.  Then

$$(a,b)_2 = (-1)^{\epsilon(u)\epsilon(v)+\alpha\omega(v)+\beta\omega(u)}.$$

In order to verify this formula, it suffices to verify it for $a,b\in S$.  Provided that $a,b\in S$ are not both divisible by 2, the equation $z^2-ax^2-by^2=0$ has a primitive solution over $\mathbb{Z}_2$ if and only if it has a primitive solution over $\mathbb{Z}/8\mathbb{Z}$.

If $a=2u$ and $b=2v$, we can use $(2u,2v)_2=(2u,2v)_2(-2v,2v)_2=(-4uv,2v)_2=(-uv,2v)_2$ to reduce to the case where $a$ and $b$ are not both divisible by 2.  Thus it becomes a finite computation over $\mathbb{Z}/8\mathbb{Z}$ to verify the formula above.