This is a follow-up to MSE question #2998091.
From that question, we were able to get the bounds $$\frac{q-1}{q(q+1)} = f(1) \leq f(k) < \frac{q-2}{q(q-1)}$$ and $$0 < f(q) \leq f(5) = \frac{(5^k - 1)(5^{k+1} - 2\cdot{5^k} + 1)}{4\cdot{5^k}(5^{k+1} - 1)},$$ where $f(q,k)$ is the function $$f(q,k) = \frac{(q^k-1)(q^{k+1}-2q^k+1)}{{q^k}(q-1)(q^{k+1}-1)},$$ defined for $q \geq 5$ and $k \geq 1$.
Now assume that $q \neq 5$ and $k \neq 1$. Solving the (resulting?) inequality $$\frac{q-1}{q(q+1)} < f(q,k) < \frac{(5^k - 1)(5^{k+1} - 2\cdot{5^k} + 1)}{4\cdot{5^k}(5^{k+1} - 1)}$$ for $q$ and $k$ I get (with some help from WolframAlpha):
$$q > 4 \text{ and } k > \frac{\log\bigg(\frac{q}{q-4}\bigg)}{\log(5)}.$$
Since $q \geq 5$ and by assumption, $q \neq 5$, the first inequality could be strengthened to $q > 5$.
How about the second inequality? $$k > \frac{\log\bigg(\frac{q}{q-4}\bigg)}{\log(5)}$$
When $q = 5$, all it says is that $k > 1$. When $q > 5$, all we could deduce is that $$\frac{4}{q} < \frac{4}{5} \implies 1 - \frac{4}{q} > \frac{1}{5} \implies \frac{q}{q-4} = \frac{1}{1-\frac{4}{q}} < 5,$$ so that we could only obtain an upper bound of $1$ for $$\frac{\log\bigg(\frac{q}{q-4}\bigg)}{\log(5)}$$ but not a lower bound.
Indeed, it appears that the inequality $$k > \frac{\log\bigg(\frac{q}{q-4}\bigg)}{\log(5)} = \log_{5}\bigg(\frac{q}{q-4}\bigg)$$ is trivial, because we have the limit $$\lim_{q \to \infty}{\log_{5}\bigg(\frac{q}{q-4}\bigg)} = 0$$ whereas we already have $k > 1$, by assumption.
Alternatively, one can perhaps get by $$\log_{5}\bigg(\frac{q}{q-4}\bigg) < k$$ through solving for $q$: $$q > \frac{4\cdot{5^k}}{5^k - 1},$$ from which we get that $q > 5$ if $k = 1$.
So perhaps the correct assumption would have been $$\lnot(q = 5 \land k = 1)?$$
Blimey, how could I have missed it!
When $f$ is considered as a function of $k$, $$f(q,1) = \frac{q-1}{q(q+1)}$$ which when evaluated at $q=5$ equals $2/15$.
When $f$ is considered as a function of $q$, $$f(5,k) = \frac{(5^k - 1)(5^{k+1} - 2\cdot{5^k} + 1)}{4\cdot{5^k}(5^{k+1} - 1)}$$ which when evaluated at $k=1$ equals $2/15$.
So indeed, in order to have an inequality in-between $$\frac{q-1}{q(q+1)} = f(q,1) \neq f(5,k) = \frac{(5^k - 1)(5^{k+1} - 2\cdot{5^k} + 1)}{4\cdot{5^k}(5^{k+1} - 1)}$$ it suffices to consider $$\frac{q-1}{q(q+1)} = f(q,1) < f(5,k) = \frac{(5^k - 1)(5^{k+1} - 2\cdot{5^k} + 1)}{4\cdot{5^k}(5^{k+1} - 1)},$$ which is then equivalent to the condition $$\lnot(q = 5 \land k = 1) \iff \lnot(q=5) \lor \lnot(k=1) \iff (q > 5) \lor (k > 1).$$