It is clear that $x=1$ is a solution of the problem (with $y$ any positive integer).

We now show that for $x>1$, the given equation has no solution. In fact, suppose that there are $x>1$ and $y$ are positive integers satisfying the equation. Then, one has $y>1$ and
$$x^{2016}+x^{2015}+\cdots+x+1 = y^{2015}-1 = (y-1)(y^{2014}+y^{2013}+\cdots+y+1).$$
Put $p=2017$, and note that $p$ is a prime. Let $q$ be a prime factor of $x^{p-1}+\cdots+x+1$. Then $(x, q)=1$ and by Fermat's little theorem we get $q \mid x^{q-1}-1$. But $q \mid x^{p}-1$, it follows that the order of $x$ modulo $q$ is a divisor of $(p, q-1)$ which is either $1$ or $p$. This shows that $q \mid x-1$ or $p \mid q-1$.

If $q \mid x-1$ then $x^{p-1}+\cdots+x+1 \equiv p \pmod{q}$, i.e. $q \mid p$, this gives $p=q$. This means that any prime divisor of $x^{p-1}+\cdots+x+1$ is congruent either to $0$ or to $1 \pmod{p}$, hence any positive divisor of $x^{p-1}+\cdots+x+1$ is too. In particular, $y-1$ and $y^{2014}+y^{2013}+\cdots+y+1$ (which are positive divisors of $x^{p-1}+\cdots+x+1$) are congruent either to $0$ or to $1 \pmod{p}$.

But, if $p \mid y-1$ then
$$y^{2014}+y^{2013}+\cdots+y+1 \equiv 2015 \not\equiv 0,1 \pmod{p},$$
we get a contradiction.

Hence, $y-1 \equiv 1 \pmod{p}$, i.e. $y \equiv 2 \pmod{p}$, then
$$y^{2014}+y^{2013}+\cdots+y+1 \equiv 2^{2014}+2^{2013}+\cdots+2+1 \equiv 2^{2015}-1 \pmod{p}.$$
This shows that $2^{2015}-1 \equiv 0 \pmod{p}$ or $2^{2015}-1 \equiv 1 \pmod{p}$. That is, either $p \mid 2^{2015}-1$ or $p \mid 2^{2014}-1$. On the other hand, by Fermat's little theorem again, $p \mid 2^{p-1}-1 = 2^{2016}-1$. From this, we easily get $p \mid 1$ or $p \mid 3$ which is impossible. The problem is therefore solved.