Why is nucleus charge squared in Rydbergs formula?
Derivation comes from the formula for electrostatic potential energy.

The electrostatic potential between the point charges of an electron, and a proton would be represented by the formula

$$U_E(r) = k_e \frac{q_eQ_p}{r}$$

However, in the derivation that we are looking at a difference of energy is from an emission or absorption, such that the energy concerned is 

$$E_i - E_f$$

The subtraction appears in the Rydberg formula from the distance in the form

$$\left( \frac {1} {n_1^2} - \frac {1} {n_2^2} \right)$$

However, the terms other than the radius are multiplied together and come outside the parentheses.  Even though we are looking at only a single electron, the term for the charge in the nucleus is squared because of this. This can be seen in the Bohr derivation of the Rydberg formula. 

$$E = h\nu = E_i - E_f = \frac {m_eq_e^2q_Z^2} {8h^2\epsilon_0} \left( \frac {1} {n_f^2} - \frac {1} {n_i^2} \right)$$

This shows the origin of the squared term for the nuclear charge which correlates with the Rydberg formula for a hydrogen-like element since all terms outside the parentheses other than charge are constant.

$$\frac {1} {\lambda_{vac}} = RZ^2\left( \frac {1} {n_1^2} - \frac {1} {n_2^2} \right)$$