This is a partial answer.

This answer proves that if $N$ is a prime, then $S_{n-2}\equiv 0\pmod N$.

*Proof* :

It can be proven by induction that
$$S_i=(2-\sqrt 3)^{2kp^{i+2}}+(2+\sqrt 3)^{2kp^{i+2}}\tag1$$

Using $(1)$ and $2\pm\sqrt 3=\bigg(\frac{\sqrt{6}\pm\sqrt 2}{2}\bigg)^2$, we get
$$\begin{align}&2^{N+1}S_{n-2}^2-2^{N+2}
\\\\&=(\sqrt 6-\sqrt 2)(\sqrt 6-\sqrt 2)^{N}+(\sqrt 6+\sqrt 2)(\sqrt 6+\sqrt 2)^{N}
\\\\&=\sqrt 6\bigg((\sqrt 6+\sqrt 2)^{N}+(\sqrt 6-\sqrt 2)^{N}\bigg)
\\&\qquad\qquad +\sqrt 2\bigg((\sqrt 6+\sqrt 2)^{N}-(\sqrt 6-\sqrt 2)^{N}\bigg)
\\\\&=\sqrt 6\sum_{i=0}^{N}\binom Ni(\sqrt 6)^{N-i}\bigg((\sqrt 2)^i+(-\sqrt 2)^i\bigg)
\\&\qquad\qquad +\sqrt 2\sum_{i=0}^{N}\binom Ni(\sqrt 6)^{N-i}\bigg((\sqrt 2)^i-(-\sqrt 2)^i\bigg)
\\\\&=\sum_{j=0}^{(N-1)/2}\binom N{2j}6^{(N+1-2j)/2}\cdot 2^{j+1}+\sum_{j=1}^{(N+1)/2}\binom N{2j-1}6^{(N-2j+1)/2}\cdot 2^{1+j}
\\\\&\equiv 6^{(N+1)/2}\cdot 2+2^{(N+3)/2}\pmod N
\\\\&\equiv 12\cdot 2^{(N-1)/2}\cdot 3^{(N-1)/2}+4\cdot 2^{(N-1)/2}\pmod N
\\\\&\equiv 12\cdot (-1)^{(N^2-1)/8}\cdot \frac{(-1)^{(N-1)/2}}{\bigg(\frac N3\bigg)}+4\cdot (-1)^{(N^2-1)/8}\pmod N
\\\\&\equiv 12\cdot 1\cdot \frac{-1}{1}+4\cdot 1\pmod N
\\\\&\equiv -8\pmod N\end{align}$$

So, we get
$$2^{N+1}S_{n-2}^2-2^{N+2}\equiv -8\pmod N$$
It follows from $2^{N-1}\equiv 1\pmod N$ that
$$S_{n-2}\equiv 0\pmod N$$