Suppose that $n = p^k m^2$ is an odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.
That $n$ is perfect essentially means that $$\sigma(p^k)\sigma(m^2)=\sigma(p^k m^2)=\sigma(n)=2n=2p^k m^2,$$ where $\sigma(x)$ is the sum of divisors of the positive integer $x$. In terms of the abundancy index $I(x) = \sigma(x)/x$, we have $$I(p^k)I(m^2)=I(p^k m^2)=2.$$ Note that both $\sigma$ and $I$ are multiplicative functions. In particular, note that when $a$ divides $b$ (denoted by $a \mid b$), then $I(a) \leq I(b)$.
Now, since $k \equiv 1 \pmod 4$, it is easy to prove (Exercise!) that $$(p+1) \mid \sigma(p^k).$$
In fact, furthermore, since $p \equiv 1 \pmod 4$, this shows that $$\frac{p+1}{2} \mid \frac{\sigma(p^k)}{2} \mid n.$$
In particular, this shows that, if $p=5$, then $3 \mid n$. This implies that $3^2 \mid m^2$.
By employing the factor chain method (see, for example, this question), we have the following: $$3^2 || m^2 \implies \sigma(3^2) = 13 \mid m^2 \implies {13}^2 \mid m^2$$ $${13}^2 || m^2 \implies \sigma({13}^2) = 3\times{61} = 183 \mid m^2 \implies {61}^2 \mid m^2$$ $${61}^2 || m^2 \implies \sigma({61}^2) = 3\times{13}\times{97} = 3783 \mid m^2 \implies {97}^2 \mid m^2$$ $${97}^2 || m^2 \implies \sigma({97}^2) = 3\times{3169} = 9507 \mid m^2 \implies {3169}^2 \mid m^2$$ $${3169}^2 || m^2 \implies \sigma({3169}^2) = 3\times{3348577} = 10045731 \mid m^2 \implies {3348577}^2 \mid m^2$$
Now, test: $$1.60723 \approx \frac{100916741461563}{62789076915421} = I({3}^2)I({13}^2)I({61}^2)I({97}^2)I({3169}^2)I({3348577}^2) \leq I(m^2) \leq \frac{2}{I(p)} = \frac{2p}{p+1} = \frac{5}{3},$$ whence there is no contradiction.
(Here are the WolframAlpha computations.)
QUESTIONS
(1) Am I doing the factor chain method properly? That is, when I encounter consequent factors which are already part of assumed factors, should I ignore the succeeding (consequent) factors? (For example, notice that I got $3$ and $13$ as consequent factors from $\sigma({61}^2)$, and that I ignored them, and only used the prime $97$, since $3$ is an assumed prime factor from the very beginning and $13$ is a consequent factor of $\sigma({3}^2)$ from the second line.)
(2) Or should I incorporate / keep track of all consequent factors in the abundancy index computations, as they occur?
