Define $b_n(x) = a_{n+1} x + a_n$ for $n \ge 0$ and any integer $x$. We have to find all $n \ge 1$ for which $\operatorname{gcd}(b_n(n), b_{n-1}(n)) > 1$. By definition, for $k \ge 1$ we have $b_{k+1}(x) = b_k(x) + b_{k-1}(x)$, hence $\operatorname{gcd}(b_{k+1}(x), b_k(x)) = \operatorname{gcd}(b_k(x), b_{k-1}(x))$ and so, using induction and substituting $x = n$,
$$\begin{align*} \operatorname{gcd}(b_n(n), b_{n-1}(n)) &= \operatorname{gcd}(b_1(n), b_0(n)) = \operatorname{gcd}(a_2 n + a_1, a_1 n + a_0) \\ &= \operatorname{gcd}(4n + 3, 3n + 1) = \operatorname{gcd}(n + 2, 3n + 1) \\ &= \operatorname{gcd}(n + 2, -5) \in \{1, 5\}. \end{align*}$$
Hence, $n a_{n+1} + a_n$ and $n a_n + a_{n-1}$ are not coprime iff $5 \mid n + 2$, and this is equivalent to $n \equiv 3 \pmod{5}$.