So $x=\left[\frac{x}{2016} \right]+2016$ $\implies x \ge \frac{x}{2016}+2016$ $\implies x \ge 2016 \times \frac{2016}{2015}$ Also $x < 2017 \times \frac{2016}{2015}$ $\implies 2016 \times \frac{2016}{2015} \le x < 2017 \times \frac{2016}{2015}$ $\implies 1 < \frac{2016}{2015} \le \frac{x}{2016} < \frac{2017}{2015} < 2$ $\implies \left[\frac{x}{2016} \right]=1$ Thus $x=1+2016=\boxed{2017}$ is the only solution.

$x=2016a+b$ where $0 \leq b<2016$ and $a$ is integer $2015a+b=2016$ $0<2015a<2016 \to a=1,b=1,x=2017$