So I know we have the Gibbs Entropy Formula $$S=k\sum_i p_i \ln p_i$$ And I've seen the probabilities for the microstates be said to be the Boltzmann probabilities $$p_i\propto e^{-\frac{E_i}{kT}}$$ But I've also seen that to obtain Boltzmann's entropy formula you assume all states have the same probability so, if $$\Omega$$ is the number of accessible microstates then we have $$S=k \ln \Omega$$ So either the probability for all microstates is the same, it is proportional to the Boltzmann Factor, or I am missing a big chunk of information.

The Boltzmann's entropy formula is valid in the microcanical ensemble, where the number of particles $$N$$, the volume $$V$$ and the energy $$E$$ are fixed. In this case we postulate that all microstates are equally probable, and define $$\Omega$$ as the number of microstates that have an energy comprised between $$E$$ and $$E + \Delta$$, where $$\Delta \ll E$$.
If we are in the canonical ensemble, in which $$N$$, $$V$$ and temperature $$T$$ are fixed, things are a bit different. In particular, the Boltzmann relation for $$S$$ is not valid, as the entropy is now given by
$$S = -\left( \frac{\partial F}{\partial T} \right)_{N,V}$$
where $$F$$ is the Helmholtz free energy. In this ensemble, the probability of a microstate, as you wrote, is proportional to the Boltzmann factor:
$$p_i\propto e^{-\frac{E_i}{kT}}$$