Confusion with the Periodic Table
So if that is the case, shouldn't period 3 have more elements, since
  it can hold up to 18 electrons, and therefore it can have up to 18
  more protons from the largest atomic number element in period 2?


Indeed, elements of the same period have the same number of electron shells, but the "problem" is that in accordance with the Madelung/Janet/Klechkowski  rule, the $\mathrm{4s}$ orbital is occupied before the $\mathrm{3d}$ orbital. As a result, the $\mathrm{3d}$ orbital could be filled only when the $\mathrm{4s}$ orbital is already filled, i.e. only for elements of the 4th period.

Similarly, the $\mathrm{5s}$, the $\mathrm{5p}$, and the $\mathrm{6s}$ orbitals are occupied before the $\mathrm{4f}$ orbital. As a result, the $\mathrm{4f}$ orbital could be filled only when the $\mathrm{5s}$, the $\mathrm{5p}$, and the $\mathrm{6s}$ orbitals are already filled, i.e. only for elements of the 6th period.