Example 10 Set $Z$ consists of $n$ elements. What is the maximum number of 3-element subsets of $Z$ such that any two of these 3-element subsets have exactly one element in common.
Let $k_{n}$ denote the required number. Suppose we select $k_{n}$ 3-element subsets from the set $Z$, such that any two of them share exactly one common element. There are three possible cases:
(1) Each element in the set $Z$ appears in at most two 3-element subsets. Let $\{a, b, c\}$ be one of the 3-element subsets, then any other 3-element subset must intersect with $\{a, b, c\}$, and at most one of the other subsets can contain element $a$, at most one can contain element $b$, and at most one can contain element $c$. Therefore, the number of subsets is at most $1 + 3 \times 1 = 4$, i.e., $k_{n} \leqslant 4$.
(2) There is an element in the set $Z$ that appears in 3 3-element subsets, but every element in the set $Z$ appears in at most 3 3-element subsets. Let $\{a, b, c\}$ be one of the 3-element subsets, then any other subset must intersect with it, and at most two of the other subsets can contain element $a$, at most two can contain element $b$, and at most two can contain element $c$. Therefore, the number of subsets is at most $1 + 3 \times 2 = 7$, i.e., $k_{n} \leqslant 7$.
(3) The set $Z$ contains an element $a$ that belongs to at least 4 3-element subsets, then the other elements in these 4 3-element subsets containing $a$ are all different, and any other subset must also contain element $a$. Otherwise, such a subset would intersect with each of the 4 3-element subsets in exactly one element, so it would contain at least 4 elements. Therefore, in this case, $1 + 2k_{n} \leqslant n$, i.e., $k_{n} \leqslant \left[\frac{n-1}{2}\right]$.

When $n=1,2,3,4,5$, it is clear that $k_{1}=k_{2}=0, k_{3}=k_{4}=1, k_{5}=2$. When $n=6$, each element in the set $Z$ appears in at most 2 3-element subsets, otherwise the union of the 3-element subsets would contain 7 elements, so case (1) holds, i.e., $k_{6} \leqslant 4$. On the other hand, let $Z=\{a, b, c, d, e, f\}$, then the 3-element subsets can be chosen as $\{a, b, c\}, \{c, d, e\}, \{e, f, a\}, \{b, d, f\}$. Thus, $k_{6}=4$.

Let $a \in \{7,8,9, \cdots, 16\}$, then when case (3) occurs, the number of 3-element subsets is at most $\left[\frac{16-1}{2}\right]=7$. When cases (1) and (2) occur, the number of 3-element subsets is also at most 7. On the other hand, if 7 elements in the set $Z$ are $a, b, c, d, e, f, g$, then there are 7 3-element subsets: $\{a, b, c\}, \{c, d, e\}, \{e, f, a\}, \{b, d, f\}, \{a, g, d\}, \{b, g, e\}, \{c, g, f\}$. Thus, when $n=7,8,9, \cdots, 16$, $k_{n}=7$.

Finally, when $n \geqslant 17$, in any case, $k_{n} \leqslant \left[\frac{n-1}{2}\right]$. Moreover, the upper bound is achieved by selecting 3-element subsets as follows: choose one element from the set $Z$ as the common element of all 3-element subsets, and pair the remaining elements (if $n$ is even, remove one), and form 3-element subsets with the common element. Therefore, when $n \geqslant 17$, $k=\left[\frac{n-1}{2}\right]$.