Given that $a_{i}$ are all positive integers, and $6^{3}=216>164$, we know that each $a_{i}$ is no greater than 5. Let $x_{1}, x_{2}, \cdots, x_{5}$ be the number of 1, 2, 3, 4, 5 in $a_{1}, a_{2}, \cdots, a_{n}$, respectively. Then,
$$\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=n \\
x_{1}+2 x_{2}+3 x_{3}+4 x_{4}+5 x_{5}=26 \\
x_{1}+4 x_{2}+9 x_{3}+16 x_{4}+25 x_{5}=62 \\
x_{1}+8 x_{2}+27 x_{3}+64 x_{4}+125 x_{5}=164
\end{array}\right.$$

where $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ are all non-negative integers.
First, determine the values of $x_{1}, x_{2}, \cdots, x_{5}$, which is equivalent to finding the non-negative integer solutions of an indeterminate system of equations.
From (4), we know $x_{5} \leqslant 1$. If $x_{5}=1$, then (3) and (4) become
$$\left\{\begin{array}{l}
x_{1}+4 x_{2}+9 x_{3}+16 x_{4}=37 \\
x_{1}+8 x_{2}+27 x_{3}+64 x_{4}=39
\end{array}\right.$$

Subtracting (5) from (6) gives
$$4 x_{2}+18 x_{3}+48 x_{4}=2$$

This cannot hold when $x_{2}, x_{3}, x_{4}$ are non-negative integers, so
$$x_{5}=0$$

Subtracting (2) from (3) gives
$$2 x_{2}+6 x_{3}+12 x_{4}=36$$

which simplifies to
$$x_{2}+3 x_{3}+6 x_{4}=18$$

Subtracting (2) from (4) gives
$$x_{2}+4 x_{3}+10 x_{4}=23$$

Subtracting (7) from (8) gives
$$x_{3}+4 x_{4}=5$$

Thus,
$$\left(x_{3}, x_{4}\right)=(5,0),(1,1)$$

Consequently, we have $\quad\left(x_{1}, x_{2}, x_{3}, x_{4}\right)=(5,3,5,0),(1,9,1,1)$.
Using the above conclusions, we get
$$\begin{aligned}
& \left(a_{1}, a_{2}, \cdots, a_{n}\right) \\
= & (1,1,1,1,1,2,2,2,3,3,3,3,3) \\
& (1,2,2,2,2,2,2,2,2,2,3,4)
\end{aligned}$$
Example 5 Find all positive integer arrays $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$, such that
$$\left\{\begin{array}{l}
a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}, \\
a_{1}+a_{2}+\cdots+a_{n}=26, \\
a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=62, \\
a_{1}^{3}+a_{2}^{3}+\cdots+a_{n}^{3}=164 .
\end{array}\right.$$