Fields having exactly one quadratic extension (up to isomorphism)
After some searches, I can answer negatively. Actually we can state more:
There are infinite fields with exactly one extension of degree $n$, for every $n \geq 1$.
(Notice that a finite field satisfies this condition, but an algebraically closed field doesn't work since has at most one extension of degree $n \geq 1$ (actually $0$ if $n \geq 2$)).
Such fields $F$ satisfy in particular $[\overline F : F] = \infty$, since there are elements of arbitrarily large degree over $F$.
A concrete example is:
The field of formal Laurent series over $\Bbb C$, denoted by $F = \Bbb C((T))$.
(Its algebraic closure is the field of Puiseux series, but we don't need that to state $[\overline F : F] = \infty$. By the way, $F$ has characteristic 0). This example has actually been discussed here. This example doesn't require the axiom of choice (the other one below does require it)...
More generally, we can look at quasi-finite fields (which also require the field to be perfect), or the even stronger notion of pseudo-finite field. An exemple is given here:
Let $Q$ be the set of all prime powers, and let $\mathcal U$ be a non-principal ultrafilter on $Q$ (i.e. for all $q \in Q$, there is $A \in \mathcal U$ such that $q \not \in A$). Then the field $$F' = \left(\prod_{q \in Q} \mathbb F_q\right)/\mathcal U$$ is a pseudo-finite field ; in particular it has exactly one extension of degree $n$, for every $n \geq 1$.
Finite fields still give loads of examples. Let $\kappa$ be any finite field, and let $q$ be any prime (not necessarily distinct from the characteristic). Now take the union $\mathcal K$ of all fields $K_n$ for which $[K_n:\kappa]=q^n$. Then $\mathcal K$ is infinite, and will have a unique extension of every degree prime to $q$, in particular only one quadratic extension, if $q\ne2$.
EDIT — Addition:
On looking more closely at your question, I see that you make some guesses about characteristic two. There the quadratic-extension picture is simultaneously simpler and more complicated. When you adjoin the square root of something, you’re making an inseparable extension of degree $p$, and the truth of the matter is that when the base is perfect, as $\overline{\Bbb F_2}$ is, and your field is finitely generated over the base, and the transcendence degree is only one, then there is precisely one inseparable extension of degree $p$, indeed only one purely inseparable extension of degree $p^m$ for each $m$. In other words, starting with $k=\overline{\Bbb F_2}(t)$, no matter what rational function you adjoin the square root of, you get the same extension, $k^{1/2}$
On the other hand, there are infinitely many nonisomorphic quadratic separable extensions of $k=\overline{\Bbb F_2}(t)$, for instance the ones gotten by adjoining the roots of $X^2+t^mX+t$.
No, it does not. Let $ F $ be an algebraic extension of $ \mathbf Q $, maximal with respect to the property of not containing any of $ \sqrt{2}, \sqrt[3]{2}, \zeta_3 \sqrt[3]{2}, \zeta_3^2 \sqrt[3]{2} $. (The existence of such an $ F $ is guaranteed by Zorn's lemma.) Then, $ F $ has precisely one quadratic extension: $ F(\sqrt{2}) $; however, $ X^{3^k} - 2 $ remains irreducible in $ F[X] $ for all $ k \geq 1 $, thus $ [\bar{\mathbf Q} : F] $ is not finite.