When does $q^b = (p^a - 1)/(p - 1)$? $\DeclareMathOperator{\Ord}{Ord}$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.

q^b = (p^a - 1)/(p - 1)
Now for q^b to be defined p is ne 1
p is an odd prime number therefore p = 3 or p = (6k - 1) or p = (6k + 1) where k is a positive integer. 
when p = 3
q^b = (3^a - 1)/(3 - 1)
when p = 6k - 1
q^b = ((6k - 1)^a - 1)/(6k - 2)
= [(6k - 1 - 1)(6n + 1)]/(6k - 2) in atleast one of the instances.

= [(6k - 2)(6n + 1)]/(6k - 2)

= 6n + 1 where n is a positive integer

when p = 6k + 1
q^b = ((6k + 1)^a - 1)/(6k)
= [(6k + 1 - 1)(6N - 1)]/(6k) in atleast one of the instances.

= [(6k)(6N - 1)]/(6k)

= 6N - 1 where N is a positive integer

