Solution:
Compare coefficients of $x^{2016}$ in the equation to obtain $a^{2016}+1=8$, i.e. $a=7^{1/2016}$. Then, take $x=2$ to obtain
$$(2 a+b)^{2016}+(4+2 c+d)^{1008}=0$$
Since the LHS is a sum of even-exponent powers, the equation will be solved in $\mathbb{R}$ if and only if both addends are zero. In particular, $b=-2 a=-2 \cdot 7^{1/2016}$. Finally, substitute these values for $a$ and $b$ in the original equation to obtain
$$\begin{aligned} 7(x-2)^{2016}+\left(x^{2}+c x+d\right)^{1008} & =8(x-2)^{2016} \\ \left(x^{2}+c x+d\right)^{1008} & =\left(x^{2}-4 x+4\right)^{1008} \end{aligned}$$
Comparing coefficients once more, we finally obtain $c=-4$ and $d=4$.