$\DeclareMathOperator{\Ord}{Ord}$When does $q^b = (p^a - 1)/(p - 1)$ for $p, q$ odd primes and $a, b$ odd integers $> 1$ ? If no examples are possible, please give a simple proof.
A proof of this for $q=3, p=5$ might be
Assume $3^b = (5^a - 1)/4$; then \begin{equation} 5^a - 4 \cdot 3^b = 1. \tag{1} \end{equation} We show that the exponents $a,b$ in (1) are even.
We see that $3^b \mid 5^a - 1$ $\Rightarrow$ $5^a \equiv 1 \bmod 3$. Also $5 \mid 4\cdot3^b+1$ or $4 \cdot 3^b \equiv -1 \bmod 5$ so that $3^b \equiv 1 \bmod 5$.
In short: \[ 3^b\equiv1 \bmod 5, \quad 5^a\equiv1 \bmod 3. \]
By a well known theorem: if $X^c \equiv 1 \bmod p$, then $\Ord(c,p) \mid c$. By inspection, $\Ord(3,5) = 4$ so that $4 \mid b$. Similarly, $\Ord(5,3) = 2$ so that $2 \mid a$.
This violates our assumption that $a,b$ are odd. Proof complete.
However, moving further:
As $a,b$ are both even we can write (1) as $1 = 5^{2A} - 4\cdot 3^{2B} = (5 - 2\cdot3^B)(5 + 2\cdot3^B)$. Thus, $5 + 2\cdot3^B = 1$ but there is no positive value of $B$ satisfying this. Therefore (1) is not true.
Note: I want to be clear that the above proof, if it is correct, is not mine but was given to me by someone I am not free to name. If it is incorrect, the fault is entirely mine.