$\DeclareMathOperator\char{char}$This question is inspired by the MSE question [Example of a non-algebraically closed field without quadratic extensions][1]. 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** every proper finite index subfield of an algebraically closed field must be **real closed** (in particular of $\char=0$) and of index $2$. Conversely every real closed field has index $2$ in its algebraic closure. This gives examples for any odd $n$ in $\char=0$.

 - Over any **finite field $\mathbb{F}_p$** by the results of [Non-algebraically closed field in which every polynomial of degree $< n$ has a root][2] 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 $\ell_1,\dotsc,\ell_k$ (potentially with multiplicity) then each of the $F_{p,\ell_i}$ is a (nonunique minimal) field that does not have any $n$-extensions. Their algebraic extensions (in particular finite ones or the algebraic closure for the other extreme) also have no $n$-extensions. This follows from the simple structure of algebraic extensions of finite fields which are in 1:1-correspondence with [Steinitz numbers][3], cf. [the book "Infinite Algebraic Extensions of Finite Fields", p. 26][4].

 - By https://mathoverflow.net/questions/16778/what-are-the-possible-sets-of-degrees-of-irreducible-polynomials-over-a-field 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](https://dx.doi.org/10.1090/pspum/008/0174555)") is currently not available to me.

These are the only examples I know off. Are there any more examples known? What fields do Gordon/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 $\char(K)=p\nmid n$) 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 [Finite field, every element is a square implies char equal 2][5] for finite fields $K^{\times}=(K^{\times})^{\ell}$ can only occur if $\char(K)=p=\ell$, 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 $\char(K)=p\nmid n$ 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 $\char(K)=p=n$. 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.


  [1]: https://math.stackexchange.com/questions/4510217/example-of-a-non-algebraically-closed-field-without-quadratic-extensions
  [2]: https://math.stackexchange.com/questions/93744/non-algebraically-closed-field-in-which-every-polynomial-of-degree-n-has-a-ro
  [3]: https://en.wikipedia.org/wiki/Supernatural_number
  [4]: http://www.ams.org/books/conm/095/
  [5]: https://math.stackexchange.com/questions/30354/finite-field-every-element-is-a-square-implies-char-equal-2