Do isomorphisms of quotients of polynomial rings by monomial ideal preserve the constants?

Question

Let $R \neq \mathbf{0}$ be a commutative ring with identity $1$ and consider the polynomial rings $S = R[x_1, \dots, x_n], T = R[y_1, \dots, y_m]$. Furthermore let $I \subseteq S, J \subseteq T$ be monic monomial ideals (i.e. $I$ and $J$ are generated by monic monimals; I don't know if this would be the correct terminology) such that $x_i \not\in I, y_j \not\in J$ for $1 \le i \le n, 1 \le j \le m$. My questions then are:

1. If $\varphi: S/I \to T/J$ is an isomorphism, $r \in R$, does $\varphi(r + I) = r' + J$ for some $r' \in R$?

2. If this does not hold, does there always exist an isomorphism $\psi: S/I \to T/J$ that has this property (given $S/I \cong T/J$)?

3. If not, what is a (as big as possible) class of rings that satisfy this property?

4. Are there any restrictions on the ideals that can make this work?

I know that this holds if $\varphi$ is also linear (by considering the module structure as a quotient module of a polynomial ring) but am not sure about the general case. My goal is to show that we must have that $n = m$.

Edit/additional info: I am mostly interested in the case of squarefree monomial ideals $I,J$ (i.e. they are generated by squarefree monomials) but thought that there might be some more generality.

Edit 2: After thinking a bit more about this I realised that I missed the obvious case of $R = \mathbb{Z}$ since every $\mathbb{Z}$-algebra homomorphism is automatically $\mathbb{Z}$-linear because it is additive.

Another class of rings I came up with is if $R$ is finite, provided $J$ is generated by squarefree monomials. To see this, assume that $\varphi(r + I) = p + J$ for some $p \in T, r \in R$ with leading term $LT(p)$ (pick your favourite monomial order) not in $J$ such that $p$ is not a constant. Then, since $J$ is squarefree, we know that for all $k \in \mathbb{N}$, we have $(p+J)^k = p^k + J$ and those are all different because the leading term of $p^k$ has degree at least $k \deg(p)$ and never lies in $J$ because $J$ is assumed to be squarefree. Thus $$ \varphi(r^k + I) = \varphi(r + I)^k = p^k + J.$$ But this means that $\varphi(r^k + I)$ takes a different value for all $k \in \mathbb{N}$, contradicting the finiteness of $R$ by injectivity of $\varphi$. Hence $\varphi(r + I) = r' + J$ for some $r' \in R$.

Edit 3: In my previous edit we actually require there to be no nilpotent elements in $R$ for the argument to work. We can also weaken the finiteness condition on $R$ to $\{ r^k | k \in \mathbb{N} \}$ is finite for all $r \in R$, the proof is still the same.