Let $q$ be prime and $k$ be a natural number. When does $\sigma(q^k)$ have a prime factor greater than $q$?
We can slightly reduce the problem by noting that
$$\sigma(q^k)=\frac{q^{k+1}-1}{q-1}$$
Thus if $p\nmid q-1$, then $p\mid \sigma(q^k)$ if and only if $p\mid q^{k+1}-1$. Clearly a prime $p>q$ does not divide $q-1$, so $\sigma(q^k)$ has a prime factor greater than $q$ if and only if $q^{k+1}-1$ does.
Using this restatement, if one takes a look at one of my answers to this question we see this is the case when $k+1\ge q$ letting $a=q$, $b=1$, and $n=k+1$ (note the question requires $b>1$, but the proof only uses Zsigmondy's Theorem in which $b>0$ suffices). However clearly this is true in more than just this case. Can we describe all cases in which this is true, or at least expand the set of $(q,k)$ for which we know that this holds?
Edit
Lets define $\kappa(n)$ to be the smallest prime congruent to $1\mod n$. A modified argument of that in the linked question, where we restrict $b$ to $1$, leads one to conclude that there exists a prime $\ge \kappa(n)$ that divides $q^n-1$. Thus if
$$\kappa(k+1)\ge q$$
(or really $\kappa(k+1)\ge q-1$ since $q\nmid q^n-1$), then $\sigma(q^k)$ satisfies the problem, slightly increasing the set of valid $(q,k)$.
