For $x = 2k$, $x^4 = 16k^4 \equiv 0 \pmod{16}$. For $x = 2k+1$, $x^4 - 1 = 8k(k+1)(2k^2 + 2k + 1) \equiv 0 \pmod{16}$, i.e. $x^4 \equiv 1 \pmod{16}$. Since $2016^3 - 1 \equiv 15 \pmod{16}$, and the sum of the numbers on the left-hand side never gives a remainder $15$ when divided by $16$, it follows that the given equation has no solution in the integers.