Assume that $|Z|\geqslant n$ but for any $x\in Z$ there exist indices $i(x)<j(x)$ such that $A_{i(x)}\triangle A_{j(x)}=\{x\}$ (where $\triangle$ stands for the symmetric difference). The graph on $\{1,2,\ldots,n\}$ with edges $(i(x),j(x))$ for all $x\in X$ contains a cycle, since the number of edges is not less than the number of vertices. But such a cycle clearly can not exist, a contradiction.

If $|Z|=n-1$, you may consider the sets $\emptyset$ and $\{1,2,\ldots,i\}$ for $i=1,2,\ldots,n-1$.