Fields having exactly one quadratic extension (up to isomorphism) 
Let $F$ be an infinite field such that $F$ has, up to field isomorphisms$^{[1]}$, exactly one extension $K/F$ of degree $2$. Does it imply  that $[\overline F : F]<\infty$ ? What happens if $F$ has characteristic $0$?

I don't think that this holds, even if characteristic $0$, but I didn't have an example of a field $F$ of characteristic $0$ ($\implies F$ is infinite) such that $[\overline F : F] = \infty$ and in $F^*$, any product of two non-squares is a square, and there is at least one non-square (see 4) below).

My thoughts:
1) The only example of such $F$ I know is $\Bbb R$: any quadratic extension embeds in $\Bbb C$, which has already degree $2$ over $\Bbb R$. My question is to know if other examples exist: if $[\overline F : F]<\infty$ then (by Artin-Schreier theorem) $F$ is a real closed field.
2) Any finite field $\Bbb F_q$ has exactly one extension of degree $n$ (namely $\Bbb F_{q^n}$), up to field isomorphisms, for every $n \geq 1$. 
3) It implies that all the quadratic extensions are isomorphic as fields, but this is not sufficient, precisely when $F$ has no quadratic extension, e.g. $\overline F=F$ (or $\bigcup_{n \geq 0} K_n$, with $K_0=\Bbb Q,K_{n+1}=\{x \in \Bbb C \mid x^2 \in K_n\}$).
4) In characteristic different from $2$, any quadratic extension of $F$ is separable and has the form $F(\sqrt a)$ where $\sqrt a \not \in F$.
Therefore, as mentioned here, the quadratic extensions of $F$ correpond to $A:=F^* / F^{*,2}$ where $F^{*,2} = \{x^2 \mid \in F^*\}$. 
Notice that $A$ is a $\Bbb F_2$-vector space, via $[a]_2 \cdot [x]_{F^{*,2}} = [x^a]_{F^{*,2}}$.
So the most interesting case is when we are looking for fields of characteristic $\neq 2$ such that $A=F^* / F^{*,2}$ has order $2$.
Equivalently, in $F^*$, any product of two non-squares is a square, and there is at least one non-square
(because $x,y$ non squares $\implies x,1/y$ non-squares $\implies x/y = a^2 \in F^{*,2} \implies [x]_{F^{*,2}} = [y]_{F^{*,2}}$).
Let $a$ be a non-square in $F^*$ and let $i = \sqrt a$. Showing that $F(i)$ is algebraically closed is not reasonable. Notice that this condition about $[F^* : F^{*,2}]=2$ is involved in 3. here, which is precisely the situation where $a=-1$ is not a square.
Then I thought to some extension $F$ of $K=\mathrm{Frac}(\Bbb R[x,y]/(x^2+y^2+1))$ since $-1$ is not a square in $K^*$ but is a sum of squares ; we just need $[F^* : F^{*,2}]=2$, and then it's a counterexample, since $F$ won't be a formally real field. I only know how to do $[F^* : F^{*,2}]=1$, see my $K_n$'s in 3).
5) I tried $F = \overline{\Bbb F_2}(t)$, because $t^{1/n}$ has degree $n$ for any $n$, so $[\overline F :F]=\infty$. I think that any extension $F\left(\sqrt{P(t)/Q(t)}\right)$ is isomorphic to $F(\sqrt t) = \overline{\Bbb F_2}(\sqrt t)$ when $P,Q \in \overline{\Bbb F_2}[t]$ i.e. $P/Q \in F$, because we have $\sqrt{a+b}=\sqrt a + \sqrt b$ in the sense that $x^2=a,y^2=b \implies (x+y)^2 = a+b$, so that $\sqrt{P(t)/Q(t)}$ is just a rational fraction in $\sqrt t$, i.e. belongs to $F(\sqrt t)$.
However, in characteristic $2$, it is not clear that all quadratic extensions arise as $F(\sqrt a)$ for some non-square $a \in F$.
6) In characteristic 0 (at least $\neq 2$), it is not clear that $F(\sqrt{t+1}) \not \cong F(\sqrt t)$. If there was a field isomorphism, then there is $u \in  F(\sqrt t)$ such that $u^2=t+1$, hence there are $a,b \in F[t]$ such that $(a(\sqrt t)/b(\sqrt t))^2 = t+1$, which yields
$a(x)^2=(x^2+1)b(x)^2$ as polynomials in $F[x]$...

$^{[1]}$ I'm only interested in fields isomorphisms, not in "field extensions" isomorphisms (i.e. not in $F$-algebras isomorphisms – these are equivalent to saying that $f : K \stackrel{\cong}{\to} K'$ commutes with the embeddings $i : F \to K$ and $i' : F \to K'$).
 A: 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 $n \geq 1$.

A: 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 $X^2+t^mX+t$.
A: 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.
A: EDIT: This is currently incorrect. I'll work on salvaging it.
Using the axiom of choice, let $s_1, s_2,\ldots$ be a sequence of numbers that are a subset of a transcendence basis of $\mathbb{C}$. Let $F$ be a maximal field extension of $\mathbb{Q}$ that doesn't contain any of the $s_i$ (Zorn's Lemma or transfinite recursion gives this). Notice that $F(s_i)$ and $F(s_j)$ are isomorphic. Now consider $K=F(s_1^2,s_2^2,\ldots)$. We now have that $K(s_i)$ and $K(s_j)$ are isomorphic and are degree $2$ extensions of $K$. It follows that $[\mathbb{C}:K]$ is infinite. One way to see this is to take the first $k$ elements of the sequence and notice that $[K(s_1,\ldots,s_k):K]=2^k$.
The issue, as mentioned in the comments, is that I need to show that $K[\sqrt{s_i^2+s_j^2}]$  doesn't mess things up.
