On "Euclidean" odd perfect numbers

In what follows, we let $$N = r^s u^2$$ be an odd perfect number given in Eulerian form, i.e. $$r$$ is the special prime satisfying $$r \equiv s \equiv 1 \pmod 4$$ and $$\gcd(r,u)=1$$. In this preprint, Brown claimed a proof for the estimate $$r < u$$. Starni, on the other hand, proved the inequality $$r < u$$ using a different method in this paper.

Say a perfect number $$M$$ is Euclidean if $$M = m_1 m_2 \cdots m_j$$, where the factors are pairwise coprime, $$j > 1$$, $$\sigma(m_i) = m_{i+1}$$ for $$i < j$$, and $$\sigma(m_j) = 2 m_1$$.

We give them this name since Euclid gave a formula for even perfect numbers of this form (with $$j = 2$$). As we know, Euler showed that all even perfect numbers have this form with $$j = 2$$.

Consider now Euclidean odd perfect numbers. It is not hard to prove the conjecture that $$r < u$$ for Euclidean odd perfect numbers in case $$j=2$$ or in case $$j > 3$$. However, the case $$j=3$$ seems hard. Say $$M = q m^2 p^{2a}$$, where the $$3$$ factors are pairwise coprime, and $$p, q$$ are primes. We might have $$\sigma(m^2) = p^{2a},$$ $$\sigma(p^{2a}) = q,$$ and $$\sigma(q) = 2m^2.$$ For such an odd perfect number $$M$$ we would have $$q > p^{2a} > m^2$$, so that $$q^2 > m^2 p^{2a}$$, which gives $$q > m p^a$$.

Hence, to prove the conjecture $$r < u$$, it would seem that one has to rule out this kind of Euclidean odd perfect number with $$j=3$$.

Here is our:

QUESTION: Do you see a way of ruling out the following system of equations? $$\begin{cases} { \sigma(m^2) = p^{2a} \\ \sigma(p^{2a}) = q \\ \sigma(q) = 2m^2 } \end{cases}$$

If my hunch is correct, one needs to concentrate on $$\sigma(p^{2a}) = q$$, to get $$\sigma(p^{2a})=\frac{p^{2a+1} - 1}{p - 1}=q. \tag{*}$$

However, I am currently unfamiliar with methods on how to solve Equation $$(*)$$.