it suffices to consider $Q(x)=x+P(x)$ then $P(x)|P(Q(x))$ hence $P(Q(x))$ is reducible,contradiction! hence $P(x)$ is constant.

not really. one of the polynomials in that product must be constant not both. taking the same example we'll obtain that $ P(P(x)+x)=cP(x) $. if $ n $ is the degree of $ P $ then $ n^{2}=n $ so $ n=0,1 $. if $ P(x)=ax+b $ then we can take $ Q(x)=ax^{2}+xv+t $ and $ v^{2}=4at+4b $ then $ P(Q(x))=(ax+\frac{v}{2})^{2} $ and we can take $ F=G $ false. so $ P $ must be constant.