Let $S(n)$ denote the set of prime factors of $n$.
For $m=1, n=2, a=2^{2}-1$ (2 is a natural number greater than 1),
$$
\begin{array}{c}
S\left(a^{m}-1\right)=S\left(2^{2}-2\right)=\{2\}, \\
S\left(a^{n}-1\right)=S\left(\left(2^{2}-1\right)^{2}-1\right)=\{2\}=S\left(a^{m}-1\right) .
\end{array}
$$

Conversely, we prove that the only $a, m, n$ satisfying $S\left(a^{m}-1\right)=S\left(a^{n}-1\right)$ are the above cases:
Let $(m, n)=d$, it is known that $\left(a^{m}-1, a^{n}-1\right)=a^{d}-1$. Therefore,
$$
S\left(a^{m}-1\right)=S\left(a^{n}-1\right)-S\left(\left(a^{m}-1, a^{n}-1\right)\right)=S\left(a^{d}-1\right) .
$$

Let $b=a^{d}, c=b-1$. Suppose $p$ is a prime factor of $\frac{n}{d}$. Since $b^{p}-1=(b-1)\left(b^{p-1}+b^{p-2}+\cdots+1\right)$, and $S\left(b^{p}-1\right) \subseteq S\left(a^{n}-1\right)=S(b-1)$, we have
$$
\left(b^{p-1}+b^{p-2}+\cdots+1, b-1\right)=(1+1+\cdots+1, b-1)=(p, b-1) \mid(b-1) .
$$

That is, $p \mid b-1$. Also, when $p>2$,
$$
\begin{aligned}
b^{p}-1 & =(1+c)^{p}-1=p c+c_{p}^{2} c^{2}+\cdots+c^{p} \\
& =p c\left(1+\frac{p-1}{2} \cdot c+c_{p}^{3} \frac{c}{p} \cdot c+\cdots+\frac{c}{p} \cdot c^{p-2}\right) . \\
\left(1+\frac{p-1}{2} \cdot c\right. & \left.+c_{p}^{3} \frac{c}{p} \cdot c+\cdots+\frac{c}{p} \cdot c^{p-2}, c\right)=(1, c)=1, \text{ which contradicts } S\left(b^{p}-1\right) \subseteq
\end{aligned}
$$
$S(b-1)=S(c)$. Therefore, $p=2$.
$S\left(b^{2}-1\right) \subseteq S(b-1)$, so the prime factors of $b+1$ are all prime factors of $b-1$. But $(b+1, b-1)=(2, b-1) \mid 2$. Therefore, $b+1=2^{l}$, where $l$ is a natural number greater than 1.

If $4 \left\lvert\, \frac{n}{d}\right.$, then $S\left(b^{4}-1\right) \subseteq S\left(a^{n}-1\right)=S(b-1)$. But $b^{4}-1=\left(b^{2}+1\right)(b+1)(b-1),\left(b^{2}+1, b-1\right)=(2, b-1)=2$, so $b^{2}+1$ must also be a power of 2, but $b^{2}+1=\left(2^{l}-1\right)^{2}+1=2^{2 l}-2^{l+1}+2$ is divisible by 2 but not by 4, which is a contradiction. Therefore, $n=2 d$. When $m>1$, $a^{d}+1=(a+1)\left(a^{d-1}-a^{d-2}+\cdots+1\right) \cdot a^{d-1}-a^{d-2}+\cdots+1$ is the sum of $d$ odd numbers, and thus is odd and greater than 1 (since $a^{d}+1>a+1$). This contradicts the left side of (1) being a power of 2.
Therefore, $d=1 . a=2^{2}-1, m=1, n=2$.
19.6. $* \star$ Find all positive integer triples $(a, m, n)$. Satisfying: $a>1, m<n$ and the set of prime factors of $a^{m}-1$ is the same as the set of prime factors of $a^{n}-1$.