Dris index at $k=5$: some prime $\ell\ne 3$ dividing $p^4+p^2+1$ divides $s$ to an odd power

Let $p$ be an odd prime and let $m,s$ be natural numbers with $s>0$ such that the first Dris relation at exponent $k=5$,

holds. Then there is a prime $\ell\neq 3$ with

In particular the Dris index $s$ is never a power of $3$, and it is never coprime to $p^4+p^2+1$ away from the prime $3$. Only the first relation is used, not $\sigma(m^2)=p^5 s$.

Idea. Write $\sigma(p^5)=(p+1)(p^2+p+1)(p^2-p+1)=2ABC$ with $A=(p+1)/2$, $B=p^2+p+1$, $C=p^2-p+1$. Then $m^2=ABC\,s$. The numbers $B,C$ are pairwise coprime, $B$ is coprime to $A$, and $\gcd(C,A)\mid 3$; moreover $B$ lies strictly between $p^2$ and $(p+1)^2$ and $C$ strictly between $(p-1)^2$ and $p^2$, so neither is a square. As $3$ divides at most one of $B,C$, one of them, $X$, is prime to $3$; it is coprime to the other factors of $ABC$, so any prime $\ell\mid X$ has $v_\ell(m^2)=v_\ell(X)+v_\ell(s)$. Since $X$ is not a square it has a prime $\ell$ with $v_\ell(X)$ odd, and then $v_\ell(s)$ is odd.

Formalization Note The statement is an elementary lemma proved for this mission; it is not taken from the literature. The hypothesis $s>0$ only excludes the degenerate solution $m=s=0$.