Powerful Sequence

We will show that, for any prime $p,$ there exists an $n$ such that $p|a_n,$ therefore showing that the only possible value of $m$ is 1.

First, we can check that $2|a_1$ and $2017|a_{2016},$ the latter due to Fermat's Little Theorem .

Now, let $p$ be a prime such that $p \neq 2, 2017.$ We claim that $p|a_{p - 2}.$

Since $p$ is neither 2 nor 2017, we can say that if $p|4034a_{p - 2},$ the $p|a_{p - 2}.$ Now, with a little algebraic manipulation and FLT, we get

$\begin{aligned} 4034a_{p - 2} &\equiv 4034(2^{p - 2} + 2017^{p - 2} + 2015 \cdot 4034^{p - 2} - 1) \\ &\equiv 2017 \cdot 2^{p - 1} + 2 \cdot 2017^{p - 1} + 2015 \cdot 4034^{p - 1} - 4034 \\ &\equiv 2017 + 2 + 2015 - 4034 \\ &\equiv 0 \!\!\!\! \pmod{p}. \end{aligned}$

Thus, $p|4034a_{p - 2},$ so $p|a_{p - 2}.$ This means that if $m$ has any prime factors, then there exists an $n$ such that $a_n$ shares at least one of those prime factors. Thus, $m$ can only be equal to $\boxed{1},$ which is the final answer.