Solution: Let $A_{1}=\{1,2,3\} . n=3, m=1$ is a solution. If $n>3$, then there is a set $A_{z}=\{1,4,5\}$ containing 1 and the 4th element. By (i), 5 is different from $1,2,3,4$.

There is also $A_{3}=\{2,4,6\}, A_{4}=\{3,4,7\}, 6,7$ are different from the previous elements.
$A_{5}=\{1,6, j\}$. By (i), $j \neq 1,2,3,4,5,6$. And by (ii), $A_{5} \cap A_{4} \neq \varnothing$, so $j=7$.
If there is an 8th element 8, then by (i) there is $A_{6}=\{1,8, t\}$, where $t \neq 2,3,4,5,6,7$. Thus $A_{6} \cap A_{4}=\varnothing$, which contradicts (ii). Therefore, $n=7$. At this point, in addition to the above $A_{1}, A_{2}, \cdots, A_{5}$, there are also $A_{6}=\{3,5, 6\}, A_{7}=\{2,5,7\}$. Thus, $m=7$.
$(m, n)=(1,3),(7,7)$ satisfy the requirements of the problem.
Another solution: Each $A$ contains three two-element subsets, so
$$
m C_{3}^{2}=C_{n}^{2} .
$$

Each $A_{i}$ containing element $j$ has two two-element subsets containing $j$. There are $n-1$ two-element subsets containing $j$ in $X$. By (i), they all belong to exactly one $A_{i}$, so there are $\frac{n-1}{2}$ $A_{i}$ containing $j$.

Consider $A_{i}$ as points, and connect each pair of points with an edge. This forms a graph with $C_{m}^{2}$ edges. By (ii), the edge between $A_{i}$ and $A_{t}$ can be labeled with the unique common element $j$ of $A_{1}$ and $A_{6}$. The edge labeled $j$ appears exactly $C_{\frac{\pi-1}{2}}^{2}$ times. Thus,
$$
C_{m}^{2}=n C_{\frac{2-1}{2}}^{2}
$$

From (1) and (2), it is not difficult to solve $(m, n)=(1,3),(7,7)$.
Note: The case $(m, n)=(7,7)$ in Example 2 is a famous "finite projective plane" in combinatorics. If the three-element subsets are considered as "lines", it can be represented by the following diagram:

However, the seventh "line" $\{2,4,6\}$ cannot be drawn as a true line on the Euclidean plane, which is a bit regrettable.
Example 2 Find all pairs of natural numbers $(m, n)$ such that the set $X=\{1,2, \cdots, n\}$ has $m$ three-element subsets $A_{1}, A_{2}, \cdots, A_{m}$ satisfying:
(i) every pair of elements of $X$ (i.e., every two-element subset) is contained in one $A_{i} (1 \leqslant i \leqslant m)$;
(ii) any two of $A_{1}, A_{2}, \cdots, A_{m}$ have exactly one common element.