$R$ with an upper bound for degrees of irreducibles in $R[x]$

Case 1: $R$ is a field

I will refer to these notes by Keith Conrad on the Artin-Schreier theorem.

We will proceed by condition on whether or not $R$ is a perfect field. If $R$ is an imperfect field then there always exist irreducible polynomials of arbitrary large degree. This follows from Lemma 2.1 in Keith's notes which tell us that if $\alpha \in F \setminus F^{p^l}$, which is non-empty because $R$ is imperfect, then $p(t)=t^{p^l}-\alpha$ is irreducible.

If $R$ is perfect then we can deduce it has finite index in its algebraic closure. Let $A$ be the algebraic closure of $R$ then if $[A:R]=\infty$ we can construct arbitrarily large extensions by adjoining elements of a basis of $A$ to $R$. Then using the primitive element theorem we can find elements of arbitrarily large degree. So if every polynomial of degree larger than $n$ is reducible then the index of $R$ in $A$ is finite.
Now if $R=A$ we are done otherwise the Artin-Schreier theorem tells us that if $1<[A:F] < \infty$ we have that $[A:F]=2$, $F$ is formally real closed and $A=F(i)$.

Case 2: R is not a field

Note that $R$ cannot be a UFD. A UFD is a field if and only if it has no prime elements. So if $R$ was a UFD that wasn't a field we could find some prime $p \in R$ and by the OP's observations we would have to have $(p)^2=(p)$. Any finitely generated idempotent ideal is generated by an idempotent, the only idempotents of an integral domain are $0$ and $1$. So $(p)=(0),R$ which is impossible, thereby $R$ cannot be a UFD if its not a field. This observation is probably not that important in the end.

Let $R$ be an integral domain such that every polynomial over $R[x]$ of degree greater than $n$ is reducible. If $K$ is the fraction field of $R$ then $K[x]$ necessarily satisfies the same condition, in particular $K$ is algebraically closed or satisfies the conclusion of Artin-Schreier. One thing to notice is that if $K$ is algebraically closed then every polynomial splits over $R$ since any $a/b \in K$ satisfies $bx-a\in R[x]$. If $K$ is not algebraically closed then every polynomial splits over $R[i]$ for much the same reason. This MO question gives some details on how to construct these sorts of rings with valuation theory.