Consider $\frac{x^{2017}-1}{x-1}=y^{2015}-1.$ Since $2017, $ a prime hence, $\frac{x^{2017}-1}{x-1}$ is the $2017th$ cyclotomic poly. Hence all the divisors of $\frac{x^{2017}-1}{x-1}$ is $0/1 \mod 2017\implies y-1 \equiv 0/1 \mod 2017 \implies y \equiv 1,2 \mod 2017.$ Clearly, for $y \equiv 1\mod 2017,$ $y^{2014}+\dots+y+1 \equiv 2015 \mod 2017.$ Not possible, hence $y\equiv 2\mod 2017 \implies y^{2014}+\dots+y+1\equiv 2^{2015}-1 \equiv 2^{-1} -1 \mod 2017.$ Now, clearly, $y^{2014}+\dots+y+1\equiv $ can't be $0 \mod 2017 \implies y^{2014}+\dots+y+1\equiv 1\mod 2017 \implies 2^{-1}\equiv 2 \mod 2017.$ Not possible. So no solutions.