Suppose such polynomial $P(x)$ exists. Let $$P(x)=a_{998}x^{998}+a_{997}x^{997}+\cdots + a_2x^2+a_1x+a_0$$ Setting this polynomial to $[P(x)]^2-1=P(x^2+1)$ and comparing corresponding coefficients give us $$a_{997}=a_{995}=\cdots =a_3= a_1 = 0$$ Therefore $P(x)$ is even. Let $$Q(x)=P(x)-\frac{1+\sqrt{5}}{2}$$ where $\frac{1+\sqrt{5}}{2}$ is a root to equation $t^2 -1=t$. Then, $Q(x)$ is also a 998-degree even function. $$\Big[Q(x)+\frac{1+\sqrt{5}}{2}\Big]^2 -1 = Q(x^2+1) + \frac{1+\sqrt{5}}{2}$$ \begin{equation} \implies Q(x)[Q(x)+(1+\sqrt{5})] = Q(x^2+1) \end{equation} Because all terms of $Q(x)$ have even power, it is possible to let $Q(x)=R(x^2)$ where $R(x)$ is a 449-degree polynomial with real coefficients. It follows that $R(x)$ has at least one real root. This means there exists a real number $w$ such that $w^2\in\mathbb{R}$ and $Q(w)=0$. Setting this to relation above yields $w^2+1\in\mathbb{R}$ is also a root to $Q(x)=0$. Therefore every element in the following infinite sequence is a root to $Q(x)$ by repeatedly substitutions. $$w^2+1 < (w^2+1)^2+1 < ((w^2+1)^2+1)^2 + 1 < \cdots$$ This cannot be possible unless $Q(x)\equiv 0$ which is contradicting to the assumption its degree is 998.