$a$ is a norm in a quadratic extension $K(\sqrt{a})/K$ if and only if $a=u^2+v^2$ for some $u,v\in K$. $K$ an arbitrary field.
Only if follows from:
If $a$ is a sum of squares then there exists an element whose norm is $-1$.
But for the other, using the equivalent condition: $-1$ is the norm of an element, I obtain that $$-1=x^2-ay^2$$ $$x^2+1=ay^2$$ for some $x,y\in K$. So if I had the conditions in the link attached, i.e. that $K$ does not contain $i=\sqrt{-1}$, then the problem would follow immediately. But for the theorem in which this shows up I'm trying to verify the proof, there are no conditions on $K$.
Edit: I think I figured out a positive solution, for $char\neq 2$
Suppose $i\in K$. Then $(a+1)^2-(a-1)^2=4a$. Then, we have that $$(\frac{a+1}{2})^2-(\frac{a-1}{2})^2=a$$ and since $i\in K$, $$(\frac{a+1}{2})^2+(i\frac{a-1}{2})^2=a.$$
So I've reduced the problem to checking for characteristic $2$.
Edit 2:
But in the field extension $\mathbb{F}_2(t)/\mathbb{F}_2(t^2)$ ($t$ a transcendental element), $t^2$ is not the sum of two squares, since that would be
$$t^2=a^2+b^2=(a+b)^2$$
Which implies that
$t\in\mathbb{F}_2(t^2)$. So we need for $char K\neq 2$.
If $i\in K$, and $K$ does not have characteristic two then every element of $K$ is a sum of two squares.
Let $a\in K$. Then there is a solution to the simultaneous equations \begin{align} x+iy&=a\\ x-iy&=1 \end{align} and then $$x^2+y^2=(x+iy)(x-iy)=a.$$