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.