Confused on Pulse Filter Bandwidth and Symbol Rate Relation
Nyquist says that you can send up to twice as many pulses (symbols) per second as the channel bandwidth $B$ with zero ISI, so you need $R_s \leq 2B$. That is all there is to it. The sinc pulse has zero excess bandwidth so the bandwidth of the signal is equal to the symbol rate, $B_s = R_s$.
About sampling, Nyquist says that you need to sample at least twice as fast as the bandwidth of the signal, so you need $F_s \geq 2B_s$, where $F_s$ is the sampling rate.
Edit
To more directly address the points in the question...

Bandwidth of pulses: the FT of a perfect rectangular pulse is a sinc function which has infinite bandwidth.

Bandwidth of zero-ISI filter: the FT of a sinc function is a perfect rectangular pulse. If the zeros in the sinc function are at $kT_s$, where $k \in (-\infty, \infty)$, then the bandwidth is $\frac{1}{T_s}=R_s$. In case the time domain signal is purely real, then the FT will be symmetric meaning that the negative frequencies give you no more information then the positive frequencies. This cuts the bandwidth in half to $\frac{R_s}{2}$.