Are there uncountably many subrings of $\mathbb Z/p\mathbb Z[x]?$ For primes $p,$ are there uncountably many subrings of $\mathbb Z/p\mathbb Z[x]?$
In this question, my answer shows that the set of subrings of $\mathbb Z[x]$ is uncountable.
Indeed, we showed that if $R$ is a commutative ring (with identity) with an ideal $I$ with $I^2\neq I,$ then there are uncountably many subrings of $R[x].$
Also, if $R$ is a commutative ring with uncountably many subrings of $R[x]$ then:

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*If $i:R\to R’$ is a ring inclusion, then $R’[x]$ has uncountably many subrings.

*If $p:R’\to R$ is an onto homomorphism, then $R’[x]$ has uncountably many subrings.

Let $n$ be the additive order of $1$ in $R.$
If $n=\infty,$ $R$ contains a subring isomorphic to $\mathbb Z,$ so $R[x]$ has uncountably many subrings.
If $n<\infty,$ then $R$ contains $\mathbb Z/n\mathbb Z.$
If $n$ is not square-free, then $\mathbb Z/n\mathbb Z$ contains a non-zero ideal $I$ such that $I^2=0,$ so again we get an uncountable set of subrings.
We’d get uncountably many subrings of $R[x]$ for any non-trivial commutative ring $R,$ if and only if we could answer “yes” to this question. But I can’t see a way to extend the argument in the original question, and I do not see any reason for it to be true.

Note: All rings here are commutative with identity, and subrings of a ring must contain the identity of the parent ring. In particular, $R_0\times\{0\}$ is not considered a subring of $R_0\times R_1$ for rings $R_0,R_1.$

Added:
Solving the prime cases also solves the square-free cases. If $\gcd(m,n)=1,$ then $\mathbb Z/(mn)\mathbb Z[x]\equiv \mathbb Z/m[x]\times \mathbb Z/n[x],$ and Chinese remainder theorem shows the subrings are all of the form $S_m\times S_n.$
For example, when $m=2,n=3,$ any $S$ gives $S_2\cong 3S\subseteq \mathbb Z/2[x]\times\{0\}$ and $S_3\cong 4S\subseteq \{0\}\times Z/ 3[x].$
 A: No, there are only countably many.
Namely, since we assume subrings contain the unit, and our base ring is $\mathbb F_p$, we are in the situation of If $k\subset R\subset k[x]$, then $R$ is Noetherian?.
The answer there implies that any such subring $S$ (called $R$ there) is generated by finitely many elements. (W.l.o.g., $S$ contains a polynomial $y$ of positive degree; then as $k[y]$-module, $S$ (being a submodule of the f.g. module $k[x]$ over the Noetherian ring $k[y]$) is finitely generated, say by $f_1, \dots, f_n$; but then $S$, as subring of $k[x]$, is generated by $y, f_1, \dots, f_n$.)
That is, there is a surjection from the set of finite subsets of our (obviously countable) polynomial ring to the set of its subrings. But the set of finite subsets of a countable set is countable.

To apply this argument, we specifically used that the coefficient field is $\mathbb F_p$, for otherwise not every subring of the polynomials necessarily contains the full coefficient ring.
However, as Ravi Fernando points out in a comment (thanks!), the argument can be generalized as follows to the situation of $K = \mathbb F_q$ a finite extension of $\mathbb F_p$. Let $S$ be a subring of $K[x]$. Still w.l.o.g., assume there is a polynomial $y \in S$ of positive degree; it generates the subring $\mathbb F_p[y]$ of $K[x]$. Now look at the inclusion chain of rings $\mathbb F_p[y] \subseteq K[y] \subseteq K[x]$; each larger ring is an f.g. module over the next smaller one (for the first inclusion because $K \vert \mathbb F_p$ is finite, and for the second it follows via the linked question as above); so we also have that $K[x]$ is a f.g. module over $\mathbb F_p[y]$. But this last ring is still Noetherian, and so our subring $S$, which as $\mathbb F_p[y]$-module is contained in $K[x]$, is finitely generated as above.
On the other hand, Thomas Andrews points out in a comment that for the case of coefficients from an infinite algebraic extension of $\mathbb F_p$, the original question is answered in the positive. (And of course it would also be for transcendental extensions.)
Summing up:
For any unital ring $R$, the polynomial ring $R[x]$ has countably many unital subrings (with the same unit) if and only if $R$ is a finite product of finite fields. Otherwise it has uncountably many.
