Let $N$ be a torsion free finitely generated nilpotent group. We define the rank of $N$ to be the natural number $m$ such that there exists a finite subnormal series $$ N=N_m \vartriangleright N_{m-1} \vartriangleright \cdots \vartriangleright N_1 \vartriangleright N_0=\{e\} $$ such that $N_j/N_{j-1} \cong \mathbb Z$ for each $1 \le j \le m$.
Can we show the rank of $N$ is exactly the minimal number of generating elements of $N$?
I don’t think so. Take the free nilpotent group of class $2$ and rank $2$, $G$, which consists of elements of the form $x^ay^b[y,x]^c$, with multiplication given by $(x^ay^b[y,x]^c)(x^ry^s[y,x]^t)= x^{a+r}y^{b+s}[y,x]^{c+t+rb}.$ I claim the rank as you define it is $3$, even though it is $2$-generated.
Indeed, if $G/N$ is isomorphic to $\mathbb{Z}$, then $N$ contains the commutator subgroup. Since $G^{\rm ab}$ is free abelian of rank $2$, then $N$ properly contains $[G,G]$, and is itself free abelian of rank $2$. Thus, it has rank $2$, so we need at least two more steps to drop down to the trivial group. This gives a “rank” as per your definition of $3$.
Replacing class $2$ with class $c$, you should be able to get arbitrarily large “rank” while still having the group be $2$-generated.