Indeed,


**Conjecture**   Let $\ p\ $ be a prime, $\ n>1\, $ -- a natural number, and let the largest prime divisor $\ q\ $ of $\ s(p^n) :=\sum\_{k=0}^n\,p^k\ $ satisfy $ q<p.\ $ Then $\ p^n\ =\ 7^3.$


The unique(?) exception would be


$$ s(7^3)\ =\ 400\ = 2^4\cdot5^2 $$



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**EDIT** A micro-observation: when primes
$\ p\ q\ r\ $ satisfy $\ q|s(p^{r-1})\ $ then
$\ p>r\ $ and $\ q\ge r$.


More generally, $\ p>\rho $ and $\ q\ge \rho\ $ when $\ \rho\ $ is the smallest prime divisor of $\ r\,\ $ where this time $\ r\ $ is **not** assumed to be a prime.



(not so) Spectacular $\ 7^3$
There's also


$$s(67^2) = 1 + 67 + 67^2 = 4557 = 3 \cdot 7^2 \cdot 31 \tag{1}\label{eq1A}$$


In this case, $q = 31 \lt 67$ meets your criteria. I haven't checked further, but I'm fairly certain there are additional exceptions.


