Smallest Prime Divisor of a Recursive Sequence

Let the sequence $(a_n)$ be defined by $$ a_1=3,\qquad a_{n+1}=a_n^2-2 $$ for every positive integer $n$. Let $p_n$ be the smallest prime divisor of $a_n$. Prove that $$ p_n\ge 2n+3 $$ for every $n\ge 2$.

Using mathematical induction, we prove that $ a_n=\alpha^{2^{n-1}}+\beta^{2^{n-1}} $ for every positive integer $n$, where $ \alpha=\dfrac{3+\sqrt{5}}{2} \text{ and } \beta=\dfrac{3-\sqrt{5}}{2}. $

Therefore, $ a_n=\lambda^{2^n}+\omega^{2^n}, $ where $ \lambda=\dfrac{1+\sqrt{5}}{2} \text{ and } \omega=\dfrac{1-\sqrt{5}}{2}. $

Since $a_n$ is odd for every positive integer $n$, $p_n$ is odd for every positive integer $n$. I have no idea how to continue. Any help is super precious