Dimension of classifying space of a group

No. Baumslag-Solitar groups of type $(1, n)$, which are semidirect products $\Bbb Z \ltimes \Bbb Z[1/n]$ have finite two-dimensional classifying spaces, but $\Bbb Z[1/n]$ clearly cannot have one-dimensional classifying space.

Famous examples come from the so-called Bestvina-Brady groups. Given a simplicial graph $\Gamma$, define the right-angled Artin group $A(\Gamma)$ as $$\langle \text{$u$ vertex of $\Gamma$} \mid [u,v]=1 \text{ if $\{u,v\}$ is an edge of $\Gamma$} \rangle.$$ Sending each generator to $1$ yields a morphism $A(\Gamma) \to \mathbb{Z}$, and its kernel $BB_\Gamma$ is referred to as a Bestvina-Brady group.

If $\Gamma$ is finite, then $A(\Gamma)$ admits a finite classifying space of dimension $\mathrm{clique}(\Gamma)$ (:= maximal cardinality of a complete subgraph in $\Gamma$), but $BB_\Gamma$ may not have good finiteness properties.

Theorem. $BB_\Gamma$ is of type $F_n$ (i.e. admits a classifying space with finite $n$-skeleton) if and only if the flag completion of $\Gamma$ is $(n-1)$-connected.

If $\Gamma$ is a pair of isolated vertices, then we recover the example mentioned by Andy Putman in the comments. Here, $BB_\Gamma$ is not even finitely generated (because $\Gamma$ is not connected). If $\Gamma$ is a $4$-cycle, then $BB_\Gamma$ is finitely generated but not finitely presented, so it does not admit a classifying space with a finite $2$-skeleton. (This example was initially due to Stallings and Bieri.) You can find examples which are of type $F_n$ but not $F_{n-1}$ for every $n \geq 2$ by considering high dimensional spheres.

However, the initial question becomes more interesting if we assume that $N$ already admits a finite classifying space:

Question: Let $G$ be a group and $N \lhd G$ a subgroup such that $G/N \simeq \mathbb{Z}$. Assume that $G$ admits a finite classifying space of dimension $d$ and that $N$ admits a finite classifying space. Is it true that $N$ admits a finite classifying space of dimension $d-1$?

No, not even in the case when $G$ has a 1-dimensional classifying space: if $G$ is free of rank at least two, then for any $N$ with $G/N\cong \mathbb{Z}$, $N$ will be free of infinite rank and so $N$ will need a 1-dimensional classifying space too.

What you get easily is that if $G$ has cohomological dimension $n$ (modulo Eilenberg-Ganea issues for $n=2$ this is the same as the minimal dimension of a classifying space for $G$) then $N$ with $G/N\cong \mathbb{Z}$ cannot have a classifying space whose dimension is lower than $n-1$. Also easily $N$ does have a classifying space of dimension $n$ provided that $G$ does. But both of these possibilities can and do occur.

Concerning the question of whether $N$ can have a finite classifying space, the Euler characteristic gives a simple obstruction. If the Euler characteristic of the classifying space for $G$, $\chi(G)$ is non-zero, then $N$ cannot have a finite classifying space, because Euler characteristics multiply for group extensions and $\chi(\mathbb{Z})=0$, giving $\chi(G)=\chi(N)\chi(\mathbb{Z})=0$ whenever $\chi(N)$ is well-defined.