If $H$ is a proper subgroup of a $p$-group $G$, then $H$ is proper in $N_G(H)$.
You are on the right track. Now look at the subgroup $H/Z$ of $G/Z$. By induction, its normalizer is strictly larger than $H/Z$. Say it contains the residue class $\overline x$ of $x \in G$ where $\overline x \not\in H/Z$. Now show that $x$ also normalizes $H$ in $G$ to get a contradiction.