On Yanqi Xu's 2016 joint undergraduate math research project with Dr. Judy Holdener at Kenyon College In what follows, we let $\sigma(X)$ denote the sum of the divisors of the positive integer $X$.  Denote the abundancy index of $X$ by $I(X)=\sigma(X)/X$, and the deficiency of $X$ by $D(X)=2X-\sigma(X)$.  Finally, let $s(X)=\sigma(X)-X$ denote the sum of the aliquot divisors of $X$.
This is a question about Yanqi Xu's 2016 joint undergraduate math research project with Dr. Judy Holdener at Kenyon College, titled Characterization of the Positive Integers with Abundancy Index of the Form $(2x-1)/x$.  (A copy of the poster presentation is available via the following hyperlink.)
In the abstract of the paper, it is stated in the fourth sentence that

Rational numbers of the form $(2x-1)/x$ are important since both even and odd perfect numbers have a divisor with abundancy index of this form.

Let $M = 2^{p-1}(2^p - 1)$ be an even perfect number, and let $N = q^k n^2$ be an odd perfect number.
Clearly,
$$I(2^{p-1}) = \frac{2^p - 1}{2^{p-1}} = \frac{2x_1 - 1}{x_1}$$
where $x_1 = 2^{p-1}$.  (In other words, $2^{p-1}$ is an even almost perfect number, since it is a power of two.)
However,
$$I(p^k) = \frac{p^{k+1} - 1}{p^{k+1} - p^k}$$
and
$$I(n^2) = \frac{2}{I(p^k)} = \frac{2(p^{k+1} - p^k)}{p^{k+1} - 1}$$
so clearly $p^k$ is not almost perfect (since $p$ must be odd).
Additionally, since
$$\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\gcd(n^2,\sigma(n^2))=\frac{D(n^2)}{s(q^k)}=\frac{2s(n^2)}{D(q^k)} \geq 3$$
(see the paper [Dris, 2012]), then clearly $n^2$ is likewise not almost perfect.  (Similarly, it can be proved that $n$ and $q^k n$ are not almost perfect.)
So I think the trivial divisor $1$ of an odd perfect number has the required abundancy index
$$I(1) = 1 = \frac{2\cdot{1} - 1}{1} = \frac{2x_2 - 1}{x_2}$$
where $x_2 = 1$.
Here is my question:

Is there any other divisor $m > 1$ of an odd perfect number $N = q^k n^2$ such that
  $$I(m) = \frac{2x - 1}{x}$$
  for some positive integer $x$?

 A: Let $N = q^k n^2$ be an odd perfect number with special/Euler prime $q$.
Suppose that the Descartes-Frenicle-Sorli Conjecture that $k=1$ holds.
Then
$$I(n^2) = \frac{2}{I(q)} = \frac{2q}{q+1} = 2 - \frac{1}{(q+1)/2} = \frac{2((q+1)/2) - 1}{(q+1)/2},$$
where $(q+1)/2$ is an integer (since $q \equiv 1 \pmod 4$).
It remains to consider the case when $k>1$.  (Note that $k \equiv 1 \pmod 4$.)
A: The following answer is based on a theorem in a paper (to appear) communicated to Dris by Holdener (co-authored with Rachfal).
Let $N = q^k n^2$ be an odd perfect number with special/Euler prime $q$.  Suppose that $k>1$.
Consider the proper factor
$$q^{\frac{k-1}{2}} n^2.$$
This has abundancy index
$$I\bigg(q^{\frac{k-1}{2}} n^2\bigg) = I\bigg(q^{\frac{k-1}{2}}\bigg)I(n^2) = I\bigg(q^{\frac{k-1}{2}}\bigg)\cdot\frac{2}{I(q^k)} = \frac{q^{\frac{k+1}{2}} - 1}{q^{\frac{k-1}{2}}(q - 1)}\cdot\frac{2q^k (q - 1)}{q^{k+1} - 1}.$$
Since $k \equiv 1 \pmod 4$, $k+1$ is even, so that
$$q^{k+1} - 1 = \bigg(q^{\frac{k+1}{2}} - 1\bigg)\cdot\bigg(q^{\frac{k+1}{2}} + 1\bigg).$$
Canceling the common factor
$$q^{\frac{k-1}{2}} (q - 1)\bigg(q^{\frac{k+1}{2}} - 1\bigg)$$
in the numerator and denominator of
$$I\bigg(q^{\frac{k-1}{2}} n^2\bigg),$$
we get
$$I\bigg(q^{\frac{k-1}{2}} n^2\bigg) = \frac{2q^{\frac{k+1}{2}}}{q^{\frac{k+1}{2}} + 1} = 2 - \frac{1}{\bigg(\frac{q^{\frac{k+1}{2}}+1}{2}\bigg)},$$
as desired.
