This is a follow-up to this question.
Question Are there rings $A$ and $B$ satisfying Conditions (a) and (b) below?
(a) The rings $A$ and $B$ are principal ideal domains, but are not fields.
(b) The spectra $\operatorname{Spec}(A[x])$ and $\operatorname{Spec}(B[x])$ are equipotent but not homeomorphic.
In this post $x$ is an indeterminate. Also, the symbols $A$ and $B$ will always denote rings satisfying (a) above.
Comments
(c) Mumford's famous drawings of $\operatorname{Spec}(\mathbb Z[x])$ and $\operatorname{Spec}(K[x,y])$ for $K$ algebraically closed can be viewed here.
(d) My motivation is to understand the topological space $\operatorname{Spec}(A[x])$ for $A$ as above. It seems to me the weakest result one can expect is that this topological space, up to homeomorphism, does not depend only on its cardinality, but, embarrassingly, I'm not event able to prove (or disprove) this, whence the above question. (Another very weak result would be that one can choose non-isomorphic $A$ and $B$ such that $\operatorname{Spec}(A[x])$ and $\operatorname{Spec}(B[x])$ are homeomorphic. I'm also unable to prove or disprove this.)
(e) Writing $|S|$ for the cardinality of the set $S$, we have $$ |\operatorname{Spec}(A[x])|=|A[x]|=|A|, $$ and the same for $B$ (see Eric Wofsey comment here). Hence the assumption in (b) can be stated as $|A|=|B|$.
(f) If $A_1$ and $A_2$ are noetherian rings (commutative with $1$), then the topological spaces $\operatorname{Spec}(A_1)$ and $\operatorname{Spec}(A_2)$ are homeomorphic if and only if they are order-isomorphic. This is because any closed subset of $\operatorname{Spec}(A_i)$ is a finite union of subsets of the form $V(\mathfrak p)$ where $\mathfrak p$ is a prime ideal.
So that we are looking for two equipotent rings $A$ and $B$ (as in (a)) such that the posets $\operatorname{Spec}(A[x])$ and $\operatorname{Spec}(B[x])$ are not isomorphic.
(g) One can describe the poset $X:=\operatorname{Spec}(A[x])$ thanks to the last proposition near of this blogpost of Qiaochu Yuan. (I tried to be slightly more explicit that Qiaochu Yuan, but I may have introduced mistakes.)
Let $P$ be a subset of $A$ such that the map $p\mapsto(p)$ is a bijection from $P$ onto the set of maximal ideals of $A$.
For each $p\in P$ let $P_p$ be the set of irreducible monic polynomials in $(A/(p))[x]$.
Let $\widetilde Q$ be the set of all $\widetilde q(x)\in A[x]$ such that $\widetilde q(x)$ is irreducible over the field of fractions of $A$ and the coefficients of $\widetilde q(x)$ generate the unit ideal of $A$.
Let $Q$ be a set of representatives of the orbit of the action of the group of units of $A$ on $\widetilde Q$.
If $p\in P$, if $f\in P_p$ and if $\widetilde f(x)\in A[x]$ is a lift of $f(x)$, then the ideal $(p,\widetilde f(x))$ does not depend on the choice of $\widetilde f(x)$. Let us denote abusively this ideal by $(p,f(x))$ (without the tilde).
For $h=1,2$ let $X_h$ be the set of prime ideals of $A[x]$ of height $h$. Then $X_1$ is the set of nonzero prime principal ideals, $X_2$ is the set of maximal ideals, and we have $X=\{(0)\}\sqcup X_1\sqcup X_2$ (in this post $\sqcup$ means "disjoint union").
Qiaochu Yuan's Proposition says that:
(h) The map $f(x)\mapsto(f(x))$ is a bijection $P\sqcup Q\to X_1$.
(i) The map sending the ordered pair $(p,f(x))$ to the ideal $(p,f(x))$ is a bijection $\bigsqcup_{p\in P}P_p\to X_2$.
We have (with obvious notation)
$\bullet\ (p)\subset(p',f(x))$ if and only if $p=p'$,
$\bullet\ (q(x))\subset(p,f(x))$ if and only if $f(x)$ divides the image of $q(x)$ in $(A/(p))[x]$.
The other inclusions are obvious.
The answer is Yes.
Indeed, if $A$ is a countable discrete valuation ring, then the set of maximal ideals of $A$ is not dense in $\operatorname{Spec}(A[x])$, but the set of maximal ideals of $\mathbb Z[x]$ is dense in $\operatorname{Spec}(\mathbb Z[x])$.
More generally:
$\bullet$ If $C$ is a discrete valuation ring, then the closure of the set of maximal ideals of $C$ is $V((p))$, where $p$ is a unifomizer of $C$.
$\bullet$ If $B$ is a Jacobson ring, then the set of maximal ideals of $B$ is dense in $\operatorname{Spec}(B)$.