$\DeclareMathOperator\char{char}$This question is inspired by the MSE question Example of a non-algebraically closed field without quadratic extensions. 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 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, cf. the book "Infinite Algebraic Extensions of Finite Fields", p. 26.

• By 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") 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 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.

Update: I finally got access to the Gordon/Straus paper. Additionally to the results presented in Gjergji Zaimi's answer the paper constructs algebraic extension fields $k$ of $\mathbb{Q}$ such that for any $n\in\mathbb{N}_{>1}\setminus{3}$ the set of all extension degrees $S(k)$ of $k$ is exactly $S(k)=\mathbb{N}_{>0}\setminus\{2,3,\ldots,n\}$. Thus for every $n\in\mathbb{N}_{>1}$ there is a field $k$ such that $n$ is part of a finite exception set that is not attained as an extension degree of $k$. These $k$ are not number fields, as they are not finite over $\mathbb{Q}$, and their construction invokes Zorn's lemma. Since there are similar results for computable fields, it seems that one should be able to get away with something weaker than the full axiom of choice.

The proof relies on $\mathbb{Q}$ having Galois extension with group $\mathfrak{A}_n$ the alternating group for every $n\in\mathbb{N}_{>0}$. Such fields $F$ with odd positive characteristic exist according to this paper of Brink and the proof should go over, although I did not check the details. For even characteristic I could not find the corresponding statement, Brink only gives a weaker result in characteristic $2$.

It seems that a complete classification of fields $k$ that have no $n$-extension is rather bold to ask for, as it is not completely classified which sets $S(k)\subset\mathcal{P}(\mathbb{N}_{>0})$ of extension can arise for any $k$.