If $q^k n^2$ is an odd perfect number with Euler prime $q$, does $I(n^2) \geq 5/3$ imply $k=1$? Let $\sigma(x)$ denote the sum of the divisors of $x \in \mathbb{N}$.  Denote the abundancy index of $y \in \mathbb{N}$ by $I(y)=\sigma(y)/y$.
If $\sigma(N)=2N$, then $N$ is said to be perfect.
Euler proved that every odd perfect number has the form $q^k n^2$ where $q$ is prime (called the Euler prime) with $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
If $k=1$, then since $q \geq 5$, then we have
$$I(q)I(n^2)=2 \Longleftrightarrow I(n^2)=\dfrac{2q}{q+1} \geq \dfrac{5}{3}.$$
If $k>1$, then we can use the bound
$$I(q^k) < \dfrac{q}{q-1} \leq \dfrac{5}{4}$$
so that
$$I(n^2)=\dfrac{2}{I(q^k)} > \dfrac{2(q-1)}{q} \geq \dfrac{8}{5}.$$
Here is my question:

If $q^k n^2$ is an odd perfect number with Euler prime $q$, does $I(n^2) \geq 5/3$ imply $k=1$?


In an answer to a related MSE question, we get that
$$I(n^2) = \dfrac{2{q^k}(q - 1)}{q^{k+1} - 1} = 2 - 2\dfrac{q^k - 1}{q^{k+1} - 1}.$$
Then $I(n^2) \geq 5/3$ implies that
$$2 - 2\dfrac{q^k - 1}{q^{k+1} - 1} \geq \dfrac{5}{3}$$
$$\dfrac{1}{6} \geq \dfrac{q^k - 1}{q^{k+1} - 1}$$
$$q^{k+1} - 6q^k + 5 \geq 0,$$
which does not force $k=1$.
$k=1$ does follow if it is known a priori that $q=5$.  In this case, $I(n^2) \geq 5/3$ and $q=5$ imply that $I(n^2)=5/3$.
UPDATE (September 03 2017 - Manila time)
Suppose that $k > 1$, so that we have $k \geq 5$ (since $k \equiv 1 \pmod 4$).  We are given that
$$\frac{2}{I(q^k)}=I(n^2) \geq \frac{5}{3}.$$
It follows that
$$I(q^k) \leq \frac{6}{5},$$
whereupon we obtain
$$I(q^5) \leq I(q^k) \leq \frac{6}{5}.$$
It follows that
$$\frac{q^6 - 1}{q^6 - q^5} = \frac{q^6 - 1}{{q^5}(q - 1)} \leq \frac{6}{5}.$$
Consequently, since $q \geq 5 > 1$, we have
$$5q^6 - 5 = 5(q^6 - 1) \leq 6(q^6 - q^5) = 6q^6 - 6q^5$$
which implies that
$$q^6 - 6q^5 + 5 \geq 0.$$
When $q=5$, we obtain
$$q^6 - 6q^5 + 5 = {q^5}(q - 6) + 5 = {5^5}\cdot(-1) + 5 < 0.$$
