Please read this first before answering. This question is only concerned with a proof of the dimension formula using the Coquand-Lombardi characterization below. If you post something that doesn't mention the characterization, then it's not an answer and is offtopic.
Background. If $R$ is a commutative ring, it is easy to prove $\dim(R[T]) \geq \dim(R)+1$, where $\dim$ denotes the Krull dimension. If $R$ is Noetherian, we have equality. Every proof I'm aware of uses quite a bit of commutative algebra and non-trivial theorems such as Krull's intersection theorem.
T. Coquand and H. Lombardi have found a surprisingly elementary characterization of the Krull dimension that does not use prime ideals at all.
T. Coquand, H. Lombardi, A Short Proof for the Krull Dimension of a Polynomial Ring, The American Mathematical Monthly, Vol. 112, No. 9 (Nov., 2005), pp. 826-829 (4 pages)
You can read the article here.
For $x \in R$ let $R_{\{x\}}$ be the localization of $R$ at the multiplicative subset $x^{\mathbb{N}} (1+xR) \subseteq R$. Then we have
$$\qquad \dim(R) = \sup_{x \in R} \left(\dim(R_{\{x\}})+1\right)\!. \label{1}\tag{$\ast$}$$
It follows that for $k \in \mathbb{N}$ we have $\dim(R) \leq k$ if and only if for all $x_0,\dotsc,x_k \in R$ there are $a_0,\dotsc,a_k \in R$ and $m_0,\ldots,m_k \in \mathbb{N}$ such that $$x_0^{m_0} (\cdots ( x_k^{m_k} (1+a_k x_k)+\cdots)+a_0 x_0)=0.$$ You can use this to define the Krull dimension.
A consequence of this is a new short proof of $\dim(K[x_1,\dotsc,x_n])=n$, where $K$ is a field. Using Noether normalization and the fact that integral extensions don't change the dimension, it follows that $\dim(R\otimes_K S)=\dim(R)+\dim(S)$ if $R,S$ are finitely generated commutative $K$-algebras. In particular $\dim(R[T])=\dim(R)+1$. This could be useful for introductory courses on algebraic geometry which don't want to waste too much time with dimension theory.
Question. Can we use the characterization \eqref{1} of the Krull dimension by Coquand-Lombardi above to prove $\dim(R[T])=\dim(R)+1$ for Noetherian commutative rings $R$?
Such a proof should not use the prime ideal characterization/definition of the Krull dimension. Notice that the claim is equivalent to $\dim(R[T]_{\{f\}}) \leq \dim(R)$ for all $f \in R[T]$.
Maybe this question is a bit naïve. I suspect that this can only work if we find a first-order property of rings which is satisfied by Noetherian rings and prove the formula for these rings. Notice that in contrast to that the Gelfand-Kirillov dimension satisfies $\mathrm{GK}\dim(R[T])=\mathrm{GK}\dim(R)+1$ for every $K$-algebra $R$.
