$R$ with an upper bound for degrees of irreducibles in $R[x]$ One very convenient property of $\mathbb{R}$ as a ring is that there is an upper bound for the degree of irreducible polynomials in $\mathbb{R}[x]$, as 

If $f\in\mathbb{R}[x]$ has degree larger than $2$, then $f$ is reducible.

However, the proof as I know depends highly on the fact that $\mathbb{C}$ is algebraically closed, and the very nice property: $z\in\mathbb{C}$ is in $\mathbb{R}$ if and only if $z=\bar{z}$.
This makes a generalization to other integral domains rather difficult. So the problem is 

For what kind of integral domain $R$, we have a finite upper bound on the degree of irreducible elements in $R[x]$?

Some most familiar examples are ruled out: in $\mathbb{Z}[x]$, $\mathbb{Q}[x]$ and $\mathbb{F}_{p}[x]$, there is not such a bound. Unfortunately these exhaust all integral domains about which I have a working knowledge.
Another result might help is Eisenstein's criterion and its generalized form, which says if we can find a prime ideal $\mathfrak{p}$ in $R$ such that $\mathfrak{p}^2\neq\mathfrak{p}$, then by picking $a\in\mathfrak{p}^2\backslash\mathfrak{p}$ we have an irreducible $a+x^d$, where $d$ can be arbitrary, and hence the upper bound is not possible.
So we only need to focus on domains where $\mathfrak{p}^2=\mathfrak{p}$ for all prime ideals. This seems to be a quite strong restriction but I am not sure what to make of it.
Can someone give a hint?
Thanks!
 A: 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. 
A: I'm not sure about integral domains in general, but the answer is pretty clear for a field $\Bbb F$:

There is a bound for the degrees of irreducible polynomials in $\Bbb F[x]$ if $\Bbb F$ has finite index in its algebraic closure.

This is the major feature that differentiates between the cases of $\Bbb R$ and $\Bbb F_p$. I'm not sure if the converse holds.

Update: You can find in the Jacob Schlather's answer (and the comments) the reasoning for why the convese must be true for perfect fields. That takes care of finite fields and rings of characteristic zero. 
