From the first equation $-1 \le x_i \le 1$, $i = 1, \ldots, 2015$, so for all $i$: $0 \le 1 + x_i \le 2$. Add both equations and get
$$x_1^{2014}(1+x_1) + x_2^{2014}(1+x_2) + \ldots + x_{2015}^{2014}(1+x_{2015}) = 0.$$
Since every item $x_i^{2014}(1+x_i) \ge 0$, then their sum equals zero if and only if every item equals zero. So all $x_i \in \{-1, 0\}$. From the first equation follows that exactly one variable is not null, from the second equation follows that this variable equals $-1$. Thus, the equation is satisfied only by such set of numbers: $x_i = -1$, $i = 1, \ldots, 2015$, $x_j = 0$, $j \ne i$.