Let $\mathbb{N}_2$ be the set of natural numbers greater than or equal to $2$. Given a field $F$ and a subset $S \subseteq \mathbb{N}_2$, I define $F$ to be an $S$-field if, for all $n \in S$, every $n$-th degree polynomial equation with coefficients in $F$ has a solution in $F$, and also, for every natural number $m \geq 2$ such that $m \notin S$, there exists at least one $m$-th degree polynomial with coefficients in $F$ which does not have a solution in $F$.
Let me give some examples. The rational numbers are a $\emptyset$-field, the complex numbers are a $\mathbb{N}_2$-field, and the real numbers are an $O$-field, where $O$ is the set of odd integers greater than or equal to $3$.
I define a subset $S$ of $\mathbb{N}_2$ to be fieldable if there exists a field $F$ such that $F$ is an $S$-field. Not every subset of $\mathbb{N}_2$ is fieldable. My question is, what are some necessary and sufficient conditions for a given subset $S$ of $\mathbb{N}_2$ to be fieldable?
I don't know the answer, but I would like to restate the question in a form that refers to $\mathbb N$ rather than $\mathbb N_2$.
Let $\mathbb N = \langle \{0,1,2,\ldots\}; +, 0\rangle$ be the additive monoid of natural numbers. For a subset $S\subseteq \mathbb N$, call a field $\mathbb F$ an $S$-field if: ($n\in S$ $\Leftrightarrow$ every polynomial in $\mathbb F[x]$ of degree $n$ has a root in $\mathbb F$).
Question. For which subsets $S\subseteq \mathbb N$ is there an $S$-field?
Observation 1. A necessary condition for the existence of an $S$-field is that the complement of $S$ ($S^c = \mathbb N\setminus S$) is a proper additive submonoid of $\mathbb N$.
Reasoning. Choose and fix an $S$-field $\mathbb F$. We want to prove that $S^c$ is a proper additive submonoid of $\mathbb N$. To show that $S^c$ is a submonoid of $\mathbb N$ we must explain why (i) $0\in S^c$ and (ii) $S^c$ is closed under addition. To show that $S^c$ is proper we must explain why (iii) $1\notin S^c$. (The proper submonoids of $\mathbb N$ are precisely the submonoids that do not contain $1$.)
For (i), we know that if $f(x)=1$, then this constant polynomial has degree zero and $f(x)=0$ has no root in $\mathbb F$, since $1\neq 0$ is part of the definition of a field. Thus $0\in S^c$.
For (ii), if $m, n\in S^c$, then there exist polynomials $f(x), g(x)\in \mathbb F[x]$ where $\deg(f)=m$, $\deg(g)=n$, and neither $f$ nor $g$ has a root in $\mathbb F$. The product $f(x)g(x)$ has degree $m+n$ and has no root in $\mathbb F$, so $m+n\in S^c$.
For (iii), every degree-one polynomial $\alpha x+\beta\in \mathbb F[x]$ has a root $-\beta/\alpha$ in $\mathbb F$, so $1\in S$, equivalently $1\notin S^c$. \\\
This is already enough to resolve a question suggested by the comments (I don't know whether every factor-closed set is "fieldable"). It is not true that every factor closed set is 'fieldable', since the set $S=\{1, 2, 4, 8\}$ is factor closed, but its complement $S^c=\{0,3,5,6,7,9,\ldots\}$ is not a submonoid of $\mathbb N$. For example, $3, 5\in S^c$, but $3+5\notin S^c$.
The necessary condition from above is not a sufficient condition. For example, it is not to hard to verify this:
Observation 2. If there is an $S$-field $\mathbb F$ and $2, 3\in S$, then $4\in S$. In particular, there is no $S$-field for $S=\{1,2,3\}$. Consequently, even though the complementary set $\mathbb N\setminus \{1,2,3\}=\{0,4,5,6,\ldots\}$ IS a proper submonoid of $\mathbb N$, there is no $S$-field such that $S^c = \mathbb N\setminus \{1,2,3\}=\{0,4,5,6,\ldots\}$.
Reasoning. Observation 2 almost follows from the quartic formula. For, if $\mathbb F$ is an $S$-field and $2, 3\in S$, then every polynomial in $\mathbb F[x]$ of degree $1$, $2$, or $3$ has a root in $\mathbb F$. If $f(x)\in \mathbb F[x]$ has degree $4$, then the quartic formula allows one to reduce the problem of finding a root of $f(x)$ to the problem of solving polynomial equations of degrees $1$, $2$, and $3$. If all linear, quadratic and cubic equations over $\mathbb F$ have solutions in $\mathbb F$, then all quartic equations over $\mathbb F$ will also have solutions in $\mathbb F$.
The above argument does not work in characteristics $2$ or $3$ (since the quartic formula involves division by $2$ and by $3$), so I will give a different argument for those characteristics. Let's maintain that $2, 3\in S$, $4\notin S$, and now assume that $\mathbb F$ is an $S$-field of characteristic $p\in \{2, 3\}$. Let $f(x)\in \mathbb F[x]$ be a quartic that has no root in $\mathbb F$.
Claim 1. $f(x)$ is irreducible in $\mathbb F[x]$.
Otherwise $f(x) = g(x)h(x)$ where the factors have smaller degree than $4$, so the factors have roots in $\mathbb F$, so $f(x)$ has a root in $\mathbb F$.
Claim 2. $\mathbb F$ is a perfect field.
Otherwise there would be some $\alpha\in \mathbb F$ such that $x^p=\alpha$ has no root in $\mathbb F$. This cannot happen, since $p=\textrm{char}(\mathbb F)\in \{2,3\}$ and every polynomial in $\mathbb F[x]$ of degree $2$ or $3$ has a root in $\mathbb F$.
Claim 3. If $\mathbb K$ is the splitting field for $f(x)$ over $\mathbb F$, then $\mathbb K$ is Galois over $\mathbb F$ and the Galois group of $\mathbb K/\mathbb F$ is a transitive subgroup of $S_4$.
This follows from Claims 1 and 2 and from Galois theory and from the fact that $\deg(f)=4$.
Claim 4. The transitive subgroups of $S_4$ are $S_4, A_4, D_4, \mathbb Z_4$ and $K$ = Klein group. In particular, any transitive subgroup of $\textrm{Gal}(\mathbb K/\mathbb F)$ has a (normal) subgroup of index $2$ or $3$.
This follows from group theory.
Claim 5. $\mathbb F$ has a field extension $\mathbb E$ such that $[\mathbb E:\mathbb F]\in \{2, 3\}$.
Let $\mathbb E$ = the fixed subfield of $\mathbb K$ for some subgroup of index $2$ or $3$ in $\textrm{Gal}(\mathbb K/\mathbb F)$.
Claim 6. There is an irreducible polynomial $m(x)\in \mathbb F[x]$ of degree $2$ or $3$.
Let $m(x)$ be the minimal polynomial over $\mathbb F$ of a primitive element of $\mathbb E$ over $\mathbb F$.
Claim 7. Contrary to Claim 6, there is no irreducible polynomial $m(x)\in\mathbb F[x]$ of degree $2$ or $3$.
Because we have assumed that every polynomial of degree $2$ or $3$ in $\mathbb F[x]$ has a root in $\mathbb F$.
Claims 6 and 7 contradict one another. The contradiction was derived from the assumption that there is a set set $S$ with $2, 3\in S$ and $4\notin S$ for which there is an $S$-field. This establishes Observation 2.