Assume that there are positive integers $a_1, a_2, \dots, a_{2017}$ and a prime $p$ as required. Then, for each $i$, there is a positive integer $k_i$ so that
$$a_i^{2017} + a_{i+1} = p^{k_i}. \qquad (*)$$
Here $a_{2018} = a_1$. The sum of all $p^{k_i}$ equals to the sum of all $a_i^{2017} + a_i$ which is even and so $p = 2$.

We claim that $a_i$ is odd for each $i$. Write $a_i = 2^{\alpha_i} \cdot (2b_i + 1)$ for some integer $\alpha_i$ and $b_i$. Since $a_i^{2017} + a_{i+1} = 2^{k_i}$ we get $2017\alpha_i = \alpha_{i+1}$ for each $i$ and so $\alpha_i = 0$.
We claim that $(a, m) = (1, 1)$ is the only solution of $a^{2017} + 1 = 2^m$. By contrary, there is a solution with $m > 1$. Then $a > 1$ and $a^{(2 \cdot 2017, 2^{m-1})} \equiv 1 \pmod{2^m}$ since $a^{2 \cdot 2017} \equiv 1 \pmod{2^m}$ and $a^{2^{m-1}} \equiv 1 \pmod{2^m}$. Thus $a^2 \equiv 1 \pmod{2^m}$ and so $a^2 - 1 \ge 2^m = a^{2017} + 1$ which is a contradiction.
The claim implies that $a_i > 1$ for each $i$. Indeed, if there is $i$ so that $a_i = 1$ then $a_{i-1} = 1$ by the claim and so $a_i = k_i = 1$ for each $i$. In this case the sum of all $k_i$ is 2018.
Thus $a_i^{2017} + a_{i+1} \ge 2^{2017} + a_{i+1}$ and so $k_i \ge 2018$. Let $k = \min(k_i)$. For each $1 \le i \le 2017$, it follows from (*) that
$$a_i^{20172017} \equiv -a_i \pmod{2^k}.$$
Since $a_i$ is odd we get $a_i^{(2(2017^{2017}-1), 2^{k-1})} \equiv 1 \pmod{2^k}$ which implies $a_i^{2^6} \equiv 1 \pmod{2^k}$. Thus $a_i^{2^6} \ge 2^{2017}$ and so $k_i \ge 30 \cdot 2017$ which means the sum of all $k_i$ is more than $2017 \cdot 2023$.