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

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

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 or $\ r\,\ $ where this time $\ r\ $ is not assumed to be a prime.

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

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

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.

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

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

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

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

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 or $\ r\,\ $ where this time $\ r\ $ is not assumed to be a prime.