Example 4 Let $n$ be a positive integer, $\varphi(n)$ denote the number of positive integers not exceeding $n$ and coprime to $n$, and $\tau(n)$ denote the number of all positive divisors of $n$. Find all positive integers $n$ that have exactly two distinct prime factors, such that
$$
\varphi(\tau(n))=\tau(\varphi(n)) .
$$
Let $n=p^{\alpha} q^{\beta}$, where $p, q$ are primes, $p<q$, and $\alpha, \beta \in \mathbf{Z}_{+}$.
$$
\begin{array}{l}
\text { Then } \varphi(n)=p^{\alpha} q^{\beta}\left(1-\frac{1}{p}\right)\left(1-\frac{1}{q}\right) \\
=p^{\alpha-1} q^{\beta-1}(p-1)(q-1), \\
\tau(n)=(\alpha+1)(\beta+1) .
\end{array}
$$

Thus, $\varphi(\tau(n))=\varphi((\alpha+1)(\beta+1))$,
$$
\tau(\varphi(n))=\tau\left(p^{\alpha-1} q^{\beta-1}(p-1)(q-1)\right) \text {, }
$$
$\varphi(\tau(n)) \leqslant(\alpha+1)(\beta+1)-1$
$$
\begin{array}{l}
=\alpha \beta+\alpha+\beta, \\
\tau(\varphi(n))=\tau\left(p^{\alpha-1} q^{\beta-1}(p-1)(q-1)\right) \\
\geqslant \tau\left(p^{\alpha-1} q^{\beta-1}(p-1)\right)+1 \\
=\alpha \beta \tau(p-1)+1 .
\end{array}
$$

If $p \geqslant 3$, then
$$
\begin{array}{l}
\tau(p-1) \geqslant 2 \\
\Rightarrow \tau(\varphi(n)) \geqslant 2 \alpha \beta+1 \\
\Rightarrow 2 \alpha \beta+1 \leqslant \alpha \beta+\alpha+\beta \\
\Rightarrow(\alpha-1)(\beta-1) \leqslant 0 \\
\Rightarrow \alpha=\beta=1 .
\end{array}
$$

Upon inspection, this does not satisfy the problem's conditions.
Therefore, $p=2$.
$$
\text { Hence } \tau(\varphi(n))=\tau\left(2^{\alpha-1} q^{\beta-1}(q-1)\right) \text {, and } q-1
$$

is even.
Let $q-1=2 m$.
Then $m$ has an odd prime factor
$$
\begin{array}{l}
\tau(\varphi(n))=\tau\left(2^{\alpha} q^{\beta-1} m\right) \geqslant 2(\alpha+1) \beta \\
\Rightarrow 2(\alpha+1) \beta \leqslant \alpha \beta+\alpha+\beta,
\end{array}
$$

which is a contradiction.
Thus, $q-1=2^{\gamma}\left(\gamma \in \mathbf{Z}_{+}\right)$,
$$
\begin{array}{l}
\tau(\varphi(n))=\tau\left(2^{\alpha+\gamma-1} q^{\beta-1}\right)=(\alpha+\gamma) \beta . \\
\text { Also, } \varphi(\tau(n))=\varphi((\alpha+1)(\beta+1)) \\
\leqslant(\alpha+1) \varphi(\beta+1) \leqslant(\alpha+1) \beta,
\end{array}
$$

Therefore, $\varphi(\tau(n))=\tau(\varphi(n))$ if and only if $\gamma=1$, i.e., $q=3$;
$\varphi(\beta+1)=\beta$, i.e., $\beta+1$ is a prime, let it be $r$; $\alpha+1$ has all prime factors dividing $\beta+1=r$, i.e., $\alpha+1=r^{t}\left(t \in \mathbf{Z}_{+}\right)$.
In summary, $n=2^{t-1} \times 3^{r-1}$ ( $r$ is a prime, $\left.t \in \mathbf{Z}_{+}\right)$.