IRL ABooklet

We start with establishing the following statement, which is valid for not necessarily different prime numbers $m$ and $n$:
$$m \mid 7^n - 3^n \quad \text{implies} \quad m = 2 \text{ or } m > n. \qquad (5)$$
First note that $m$ cannot be equal to $3$ or $7$, hence there exists an integer $a$ such that $3a \equiv 7 \pmod m$. The assumption $3^n \equiv 7^n \pmod m$ implies $a^n \equiv 1 \pmod m$, hence $\text{ord}_m(a) \mid n$. Because $n$ is prime, we can only have $\text{ord}_m(a) = 1$ or $\text{ord}_m(a) = n$. When $\text{ord}_m(a) = 1$ we have $3 \equiv 7 \pmod m$, which is only possible when $m = 2$. When $\text{ord}_m(a) = n$ we recall that we know from Little Fermat that $\text{ord}_m(a) \le m-1$. This gives $n < m$ as desired.

Returning to our problem, we now see that $p$ must divide $7^p - 3^p$, because $p \le q \le r$. It follows from (5) that this then is only possible if $p = 2$. Because $(7^2 - 3^2)/2 = 2^2 \cdot 5$, we obtain
$$qr \mid 2^2 \cdot 5 \cdot (7^q - 3^q)(7^r - 3^r).$$
As $q \le r$, by (5) we must have $q \mid 2 \cdot 5 \cdot (7^q - 3^q)$ and so $q \in \{2, 5\}$, by (5) again.

When $p = q = 2$, we have $r \mid 2 \cdot 5 \cdot (7^r - 3^r)$ and so $r \in \{2, 5\}$ by (5).

When $p = 2$ and $q = 5$, we have to calculate $7^5 - 3^5$. Without a calculator, this can be easily done:
$$7^5 = 49 \cdot 343 = 50 \cdot 343 - 343 = 17150 - 343 = 16807.$$
$3^5 = 9 \cdot 9 \cdot 3 = 81 \cdot 3 = 243$ and $7^5 - 3^5 = 16807 - 243 = 16564$.

Finally we factorise easily $16564 = 4 \cdot 4141 = 2^2 \cdot 41 \cdot 101$. This shows that
$$r | 2 \cdot 41 \cdot 101 \cdot (7^r - 3^r).$$
From (5) we obtain now that $r \in \{2, 41, 101\}$. But $5 = q \le r$ and we are left with $r \in \{41, 101\}$ if $p = 2$ and $q = 5$.

This shows that a complete list of solutions $(p, q, r)$ is
$$(2, 2, 2), \ (2, 2, 5), \ (2, 5, 41), \ (2, 5, 101).$$