Background
There are various rational functions such that
$$\sum_{n=2}^{\infty} f(n) \left( \zeta(an)-1 \right) = q \label{1} \tag{1}$$
for some $a, q \in \mathbb{Q} \setminus \{ 0 \}$ and $f: \mathbb{N} \to \mathbb{Q} \setminus \{0\}$.
For instance, we have the following rational zeta series:
$$ \sum_{n=2}^{\infty} \left( \zeta(n)-1 \right) = 1. \label{2} \tag{2} $$
Moreover, we have:
$$ \sum_{n=1}^{\infty} \left(\zeta(2n)-1 \right) = \frac{3}{4} \tag{3}\label{3}, $$ and
$$ \sum_{n=1}^{\infty} \frac{\zeta(2n)}{2^{2n-1}} = 1, \tag{4}\label{4}$$
and finally
$$ \sum_{n=1}^{\infty} \frac{\zeta(2n)-1}{2^{2n}} = \frac{1}{6} \ .\tag{5}\label{5} $$
Question
I wonder whether there are also functions such that
$$\sum_{n=2}^{\infty} f(n) \left( \zeta(an)-1 \right)^{2} = q \tag{*}\label{*}$$
for some $a, q \in \mathbb{Q} \setminus \{ 0 \}$ and $f: \mathbb{N} \to \mathbb{Q} \setminus \{0\}$.
The series for which $f$ is the identity function and $a=1$ has - to the best of my knowledge - not been evaluated yet. Some efforts to obtain a closed form can be found in this question, this one , and over here. So it might be hard to evaluate similar series. However, I thought it might become easier if one is able to somehow pick a cleverly chosen function $f(\cdot)$.
Note: I have also asked this question on MSE some months ago
