Example 9 (1999 Canadian Mathematical Olympiad) Determine all positive integers $n$, satisfying $n=$ $[d(n)]^{2}$, where $d(n)$ denotes the number of all positive divisors of $n$ (including 1 and itself).
$d(n)=\left(1+\alpha_{1}\right)\left(1+\alpha_{2}\right) \cdots\left(1+\alpha_{k}\right)$.
By the problem, $n=[d(n)]^{2}$, then $n$ is a perfect square, $\alpha_{i}$ is even, let $\alpha_{i}=2 \beta_{i}$, thus $p_{1}^{2 \beta_{1}} p_{2}^{2 \beta_{2}} \cdots p_{k}^{2 \beta_{k}}=\left(1+2 \beta_{1}\right)^{2}\left(1+2 \beta_{2}\right)^{2} \cdots\left(1+2 \beta_{k}\right)^{2}$, that is $p_{1}^{\beta_{1}^{2}} p_{2}^{3} \cdots p_{k}^{k_{k}}=\left(1+2 \beta_{1}\right)\left(1+2 \beta_{2}\right) \cdots\left(1+2 \beta_{k}\right)$.
The right side of the above equation is odd, so $p_{i}$ is an odd prime, $p_{i} \geqslant 3$.
When $\beta_{i}>0$, we have $p_{i}^{\beta_{i}} \geqslant 3^{\beta_{i}}=(1+2)^{\beta_{i}}>1+2 \beta_{i}$, $p_{1}^{\beta_{1}^{2}} p_{2}^{\beta} \cdots p_{k}^{\beta_{k}}>\left(1+2 \beta_{1}\right)\left(1+2 \beta_{2}\right) \cdots\left(1+2 \beta_{k}\right)$.
$(* *)$
Equation (*) contradicts equation (**).
Therefore, for all $i$, $\beta_{i}=0$. Hence $n=1$.