It is known that, for $q$ prime and $k$ a positive integer,
$$I(q^k) = \frac{\sigma(q^k)}{q^k} = \frac{1 + q + \ldots + q^k}{q^k} = \frac{q^{k+1} - 1}{q^k (q - 1)},$$
where $\sigma(x)=\sigma_1(x)$ is the classical sum of divisors of the positive integer $x$.
This then gives the (trivial) upper bound
$$I(q^k) < \frac{q^{k+1}}{q^k (q - 1)} = \frac{q}{q - 1}.$$
We prove the following theorem here.
THEOREM
If $q$ is a prime number and $k$ is a positive integer, then
$$I(q^k) < \bigg(\frac{q}{q - 1}\bigg)\bigg(\frac{q^{k+1}}{q^{k+1} + 1}\bigg).$$
Moreover, we have
$$\bigg(\frac{q}{q - 1}\bigg)\bigg(\frac{q^{k+1}}{q^{k+1} + 1}\bigg) < \frac{q}{q - 1}.$$
PROOF
Let $q$ be a prime number, and let $k$ be a positive integer.
First, we want to show that
$$I(q^k) < \bigg(\frac{q}{q - 1}\bigg)\bigg(\frac{q^{k+1}}{q^{k+1} + 1}\bigg).$$
Assume to the contrary that
$$\frac{q^{k+1} - 1}{q^k (q - 1)} = I(q^k) \geq \bigg(\frac{q}{q - 1}\bigg)\bigg(\frac{q^{k+1}}{q^{k+1} + 1}\bigg).$$
This implies that
$$q^{2k+2} - 1 = (q^{k+1} - 1)(q^{k+1} + 1) \geq {q^k}\cdot{q}\cdot{q^{k+1}} = q^{2k+2},$$
which is a contradiction. This proves the first part of the proposition.
Next, we have to prove that
$$\bigg(\frac{q}{q - 1}\bigg)\bigg(\frac{q^{k+1}}{q^{k+1} + 1}\bigg) < \frac{q}{q - 1}.$$
Suppose to the contrary that
$$\bigg(\frac{q}{q - 1}\bigg)\bigg(\frac{q^{k+1}}{q^{k+1} + 1}\bigg) \geq \frac{q}{q - 1}.$$
It follows that
$$q^{k+1} \geq q^{k+1} + 1,$$
which is a contradiction. This proves the second part of the proposition, and we are done.
