sequences and series - Evaluating $sum_{n=1}^{infty} frac{phi(n)}{7^n + 1}$, where $phi(n)$ is Euler's totient function

Evaluate

$$\sum_{n=1}^{\infty} \frac{\phi(n)}{7^n + 1}$$
where $\phi(n)$ is Euler's totient function.

I found this problem on a Discord server I just joined. The first time I saw this sum, I was daunted. After gathering courage, I started my work on this problem.

I tried various things, like using the definition

$$\phi(n) = n\left(1 - \frac{1}{p_1}\right)\cdots\left(1-\frac{1}{p_m}\right)$$ (where the $p_i$ are the prime factors of $n$) and what not, but it lead to nothing.

Then I started thinking about the $7$ in the denominator; why only $7$? What is so special about $7$? What is the difference between

$$\sum_{n=1}^{\infty}\frac{\phi(n)}{6^n + 1}$$ and
$$\sum_{n=1}^{\infty}\frac{\phi(n)}{7^n + 1}$$ .
To answer that question, I took the literal difference of the two sums, but the denominator blew up so I started to think again. Because of that, I defined the two sequences and their sum:
$$a_n = \frac{\phi(n)}{7^n + 1} \textrm{ and }b_n= \frac{\phi(n)}{7^n - 1}$$
$$S_1 = \sum_{n=1}^{\infty} a_n\textrm{ and }S_2 = \sum_{n=1}^{\infty} b_n$$
and took THEIR difference, which lead to

$$\sum_{n=1}^{\infty}(b_n - a_n) = 2\sum_{n=1}^{\infty}\frac{a_nb_n}{\phi(n)}$$

Now I can say that I'm genuinely stuck. Can someone provide a hint as to what to do next?