This question is inspired by this MSE question. To repeat:

Given $n\in\mathbb{N}_{>1}$ which fields $F$ do not have extensions of degree $n$?

For short we say: If $F$ does not admit any extensions of degree $n$ than "$F$ has no $n$-extensions".

There are some results and posts about this, here is a summary of what I could find so far:

• Clearly all algebraically closed fields $F$ have no $n$-extensions for all $n\in\mathbb{N}_{>1}$.

• By the Artin-Schreier theorem any proper finite index subfield of an algebraically closed field must be real closed (in particular of $\text{char}=0$) and of index $2$. Conversely any real closed field has index $2$ in its algebraic closure. This gives examples for any odd $n$ in $\text{char}=0$.

• Over any finite field $\mathbb{F}_p$ by the results of this post one can, for any prime $n=\ell$, construct (minimal) extensions $F_{p,\ell}=\bigcup_{m\in\mathbb{N}}\mathbb{F}_{p^{(\ell^m)}}$ that do not have $\ell$-extensions. If $n$ has prime factors $\ell_1,\ldots,\ell_k$ (potentially with multiplicity) then the composite field $F_{p,n}$ of the $F_{p,\ell_i}$ is a (minimal) field that does not have any $n$-extensions. Algebraic extensions of $F_{p,n}$ (in particular finite ones or the algebraic closure for the other extreme) also have no $n$-extensions.

• By this post a similar construction as for finite fields works for the field $\mathbb{C}((t))$. It also says more about fields with prescribed sets of $n$, but does not comment on the fields and the reference of the post (B. Gordon and E.G. Straus, "On the degrees of finite extensions of a field") is currently not available to me.

These are the only examples I know off. Are there any more examples known? What fields do Gorden/Straus consider? Are there any results about the properties of such extensions? Are there any candidates whose status is unclear at the moment? Is there a complete classification of such fields?

Here are some results that might be useful in this question:

By Kummer theory any field $K$ that contains $n$ distinct $n$-th roots of unity (necessarily then $\text{char}(K)=p\nmid n$) has its abelian extensions of exponent dividing $n$ classified by $K^{\times}/(K^{\times})^n$. In particular if we have a prime $n=\ell\neq p$ any normal degree $\ell$-extension is abelian and if $K$ has no $\ell$-extension then necessarily $K^{\times}=(K^{\times})^\ell$. So it would be nice to know, which fields have this property for their multiplicative group.
Generalizing this post for finite fields $K^{\times}=(K^{\times})^{\ell}$ can only occur if $\text{char}(K)=p=\ell$, this unfortunately is exactly when Kummer theory will never apply - however the finite field case is done by the examples above.
This leaves non-finite fields: Which fields $K$ (if any at all) have $K^{\times}=(K^{\times})^n$? Do some of them have $\text{char}(K)=p\nmid n$ and include all $n$-th roots of unity? What about their non-normal degree $n$ extensions? (Kummer theory can only deal with normal extensions, but for $n>2$ there usually are also non-normal extensions.)

Artin-Schreier theory fills the gap of Kummer theory for $\text{char}(K)=p=n$. Can this be of any use to construct examples or prove their nonexistence? Assume that $K$ is transcendental over its prime field. Next to extensions that are normal and separable, there are even inseparable extensions to deal with in this case.

Disclaimer: When speaking of "the" algebraic closure, "the" composite field etc. I always assumed relative to some appropriate fixed algebraically closed overfield.