If $q^k n^2$ is an odd perfect number with special prime $q$, then can $n^2 - q^k$ be a cube? Let $q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$.
This question is an offshoot of the following earlier post:
If $q^k n^2$ is an odd perfect number with special prime $q$, then $n^2 - q^k$ is not a square.
My question in the present post is the following:

If $q^k n^2$ is an odd perfect number with special prime $q$, then can $n^2 - q^k$ be a cube?

Not much can be said without some additional hypotheses.
In the paper titled Improving the Chen and Chen result for odd perfect numbers (Lemma 8, page 7), Broughan et al. show that if
$$\frac{\sigma(n^2)}{q^k}$$
is a square, where $\sigma(x)$ is the sum of divisors of $x \in \mathbb{N}$ and $q^k n^2$ is an odd perfect number with special/Euler prime $q$, then $k=1$.  It follows that, if $\sigma(n^2)/q^k$ is a square, then $\sigma(n^2) \equiv 1 \pmod 4$, which implies that $q \equiv k = 1 \pmod 8$.
Since $n$ is odd, $n^2 \equiv 1 \pmod 8$.  This implies that $8 \mid (n^2 - q^k)$.
Alas, this is where I get stuck.  I do not even know how to rule out the simplest case
$$n^2 - q^k = 8.$$
An Elementary Attempt to Rule Out $n^2 - q^k = 8$
Subtract $9$ from both sides, and transfer $q^k$ to the RHS:
$$n^2 - 9 = q^k - 1.$$
Factoring both sides, we get
$$(n+3)(n-3)=(q-1)Q$$
where $Q=\sigma(q^{k-1})$.
We get two cases:
Case 1 $n+3 \mid (q^k - 1)$
This implies that $n + 3 \leq q^k - 1$, from which we obtain
$$n < n + 4 \leq q^k.$$
But we know from work of Brown (2016), Dris (2017), and Starni (2018) that $q < n$ holds unconditionally.
Thus, we get
$$q < n < q^k$$
which contradicts $k=1$.
Case 2 $n - 3 \mid (q^k - 1)$
This implies that
$$n - 3 \leq q^k - 1$$
which means that
$$n - 2 \leq q^k.$$
Again, by the results of Brown, Dris, and Starni, we have
$$q - 2 < n - 2 \leq q^k,$$
which does not contradict $k=1$.

Can anything be said about the general case, preferably by relaxing the hypotheses?

UPDATE (1 NOV 2019 - 08:32 AM Manila time)
Blimey!  I did not realize that
Case 1 $n+3 \mid (q^k - 1)$
holds in general.
Thus, under the assumption that $\sigma(n^2)/q^k$ is a square, $n^2 - q^k \neq 8$.
 A: Let $q^k n^2$ be an odd perfect number (OPN) with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Suppose to the contrary that $n^2 - q^k = t^3$, for some positive integer $t$.  Since $n^2 - q^k \equiv 0 \pmod 4$, then $t \geq 2$.
Substituting
$$y=n$$
$$t=x$$
the condition $n^2 - q^k = t^3$ gives rise to the Mordell curve
$$E: y^2 = x^3 + q^k$$
which has discriminant
$$D_E = -27{q^{2k}}.$$
By the Nagell-Lutz Theorem, we need to consider the possibilities
$$y \in \{\pm 1, \pm 3, \pm q, \pm 3q, \ldots, \pm q^k, \pm 3q^k\}.$$
Since $\gcd(y,q)=\gcd(n,q)=1$, then
$$y \not\in \{\pm q, \pm 3q, \ldots, \pm q^k, \pm 3q^k\}.$$
Additionally, since $q$ is a prime satisfying $q \equiv 1 \pmod 4$, then $q \geq 5$.  By work of Brown (2016), Dris (2017), and Starni (2018), we know that $n > q$.  It follows that
$$y = n > q \geq 5,$$
so that we have
$$y \not\in \{\pm 1, \pm 3\}.$$

We therefore conclude that the elliptic curve
$$E : y^2 = x^3 + q^k$$
does not have any lattice points satisfying the OPN constraints.  It follows that the equation
$$n^2 - q^k = t^3$$
does not have any positive integral solutions, if $q^k n^2$ is an OPN with special prime $q$.

In particular, this appears to settle Conjecture 6.2 and Conjecture 6.3 on page 6 of this preprint (Dris, San Diego (2020)).
Edit (July 31, 2020 - 9:44 AM Manila time) It remains to consider the interesting non-torsion integer points.
