Dris index $s \\ge 2$ odd is impossible when $k+1$ has at least two odd prime factors

Let $p$ be a prime, let $k \ge 1$ be such that $k+1$ has at least two odd prime divisors, let $m$ be odd with $p \nmid m$, and let $s \ge 2$ be odd with $s \mid m^2$. Then the Dris relations

cannot both hold; equivalently, there is no odd perfect number $N = p^k m^2$ in Euler form whose Dris index $s = 2m^2/\sigma(p^k)$ is at least $2$, for such an exponent $k$.

This is the complement, in the special exponent $k$, of the case already isolated on the platform: when $k+1$ has at most one odd prime divisor, an index $s$ that is $1$ or an odd prime is ruled out by known results, and the remaining composite case is recorded as OddPerfectNumber.no_dris_one_odd_prime_core. The present statement is the residual core for all other exponents, that is, whenever $k+1$ carries two or more distinct odd primes; it generalises the exponent-specific core OddPerfectNumber.no_dris_thirteen_core from $k \ge 13$ with $p \equiv k \equiv 1 \pmod 4$ to an arbitrary prime $p$ and an arbitrary $k \ge 1$.

The difficulty is that the standard counting bound for the prime support of $m$, namely $\omega(m) \le \omega(s) + \Omega(s) + \#\{\text{odd primes of } k+1\}$, becomes weaker precisely when $k+1$ has several odd prime factors, so Sylvester's bound $\omega(N) \ge 5$ no longer yields a contradiction.