If $A^m = 0$, then $\mbox{rank}(A) \leq \frac{m-1}{m}{n}$

I would rather prove this quite differently.

Below is the Sylvester's Rank Inequality: $$r(A)+r(B) \le n + r(AB)$$ It could be generalized by induction as: $$\sum_{k=1}^m r(A_k) \le n(m-1) + r\left(\prod_{k=1}^m A_k\right)$$

Taking $A_1=A_2=...=A_m = A$, we will get $$m\cdot r(A) \le n(m-1) + r(A^m) \implies \boxed{r(A) \le \frac{m-1}{m}n}$$ as desired

This is essentially correct. You should clarify that by $B=\left.A\right\rvert_{\operatorname{im}A}$ you mean the linear map $\operatorname{im}A\to\operatorname{im} A$ and not the map $\operatorname{im}A\to\Bbb R^n$. Depending on how much the person whom you are speaking to values your ability to fill in details, you might want to address the fact that:

1. your inductive hypothesis is that the inequality holds $(\forall k<m,\forall n,\forall A,\cdots)$ as opposed to, say, $n$ being fixed.

2. the inductive hypothesis speaks of matrices, but you have chosen the notation in the inductive step to speak of linear maps to a substantial length. This may be addressed easily in three ways that I can think of: (a) saying that in this specific instance the problem may be reformulated matricially; (b) describing the matrix of $B$ in some basis; (c) starting all over again by saying that you are considering the linear-map formulation of the problem.