If positive integer $n$ is not power two and $n^2+n+1 \mid 4^n+2^n+1$, then what sequence of $n$?
Source problem from Prove that there are infinitely many integers $n>0$ such that $n^2+n+1$ divides $4^n+2^n+1$.
We have two first term of sequence: 215,3692374808.
Let $m=n^2+n+1$. For both first terms $m$ is prime.
Question 1: can be $m$ is not prime for next terms?
Suppose 1: $m$ is prime $\overset{FLT}{\implies} 2^{n(n+1)}\equiv 1 \pmod{m} \implies 2^{3n}\equiv 1 \pmod{m}$.
Question 2: how prove if $m$ is prime then $n\equiv2\pmod{3}$?
Note: from post Modular arithmetic with Legendre symbol implies $n\not\equiv1\pmod3$.
Suppose 2: $n\equiv2\pmod{3} \implies 2^{\frac{n(n+1)}{3}}\equiv3^{\frac{n(n+1)}{3}}\equiv n^x\pmod{m}$, where some integer $x\ge0$.
Question 3: what other conditions for $n$ we can seen for speedup find next terms?