I just read the idea of Euler's proof that $\zeta(2) = \frac{\pi^2}{6}$. He studied the Taylor series of the function $\frac{\sin \pi x}{x}$.
Does someone know if he by chance was studying this function and found the connection with the Riemann Zeta Function? Or did he have some intuition that this was the right function to work with?
Obligatory "not a full answer, but too long for comments". That being said, this will get you 90% of the way there.
As mentioned in my comment, the idea of the proof is essentially covered in Mathologer's video on "Euler's crazy pi formula generator". The difference is we start by looking at Euler's product formula for $\sin x$, which can be manipulated or derived to/from many other series', such as a Taylor series for $\sin(\frac{\pi x}{x})$.
$$\sin x = x\left(1-\frac{x}{\pi}\right)\left(1+\frac{x}{\pi}\right)\left(1-\frac{x}{2\pi}\right)\left(1+\frac{x}{2\pi}\right)\left(1-\frac{x}{3\pi}\right)(\dots)$$
This is the foundation for exploring it's relationship to the zeta function. We begin by looking at pairs of terms, and notice they can be simplified as they bare the of the form of the difference of two squares:
$$\sin x = x\left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{(2\pi)^2}\right)\left(1-\frac{x^2}{(3\pi)^2}\right)(\dots)$$
The next step is a bit tricky, and I must admit I don't entirely grasp it (again, better described by Mathologer). The gist of it is that we now essentially get clever with our choices in expanding and grouping specific terms, which yields:
$$\sin x = x+\left(-\frac{1}{\pi^2}-\frac{1}{(2\pi)^2}-\frac{1}{(3\pi)^2}-\dots\right)x^3+(\dots)x^5+(\dots)$$
Now, after differentiating thrice, we're left with a series preceded by a very useful $\color{red}{\text{constant}}$.
$$-\cos x = \color{red}{\left(-\frac{1}{\pi^2}-\frac{1}{(2\pi)^2}-\frac{1}{(3\pi)^2}-\dots\right)6}+(\dots)60x^2+(\dots)$$
Evaluating this at $x=0$ gets rid of all the terms proceeding the constant.
$$\cos 0 =1= \left(\frac{1}{\pi^2}+\frac{1}{(2\pi)^2}+\frac{1}{(3\pi)^2}+\dots\right)6$$ Which can trivially be rearranged to find a formula for $\frac{\pi^2}{6}$, which just-so-happens to be of the same form as $\zeta(2)$. $$\frac{\pi^2}{6}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\dots=\zeta(2)$$
QED
As for if Euler knew beforehand that this process would result in a zeta-function-like form, is unclear. As with all long term mathematicians, both inexplicable intuition (as with the case of a lot of Euler's work) and stumbling-upon-by-chance is inevitable.
Note: I imagine it to be very difficult to algebraically manipulate these formulas for other values of the zeta function. Hence, I don't believe studying this series for sole the intention of it's general relationship to the zeta function was his intention. Clearly, it's very much related to $\zeta(2)$ and, as I imagine, also consequently related to $\zeta(2k)$ (as those are the values for which $\zeta$ yields multiples of powers of $\pi$).