Let's determine the polynomial $q(x)$ such that the polynomial $p(x)=x^{2}+x+1$ satisfies the identity $p^{2}(x)-2 p(x) q(x)+q^{2}(x)-4 p(x)+3 q(x)+3 \equiv 0$.

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Determine the polynomial $q(x)$ such that the polynomial $p(x)=x^{2}+x+1$ satisfies the identity $p^{2}(x)-2 p(x) q(x)+q^{2}(x)-4 p(x)+3 q(x)+3 \equiv 0$.
The condition of the problem can be rearranged into the quadratic equation in $q$: $q^{2}-(2 p-3) q+\left(p^{2}-4 p+3\right) \equiv 0$. After substituting $p(x)=x^{2}+x+1$, we obtain two solutions: the polynomials $q_{1}=x^{2}+2 x$ and $q_{2}=x^{2}-1$. Using these, the left side of the identity required by the problem can be factored as follows:

$$
\begin{gathered}
p^{2}(x)-2 p(x) q(x)+q^{2}(x)-4 p(x)+3 q(x)+3 \equiv \\
\equiv q^{2}(x)-\left(2 x^{2}+2 x-1\right) q(x)+\left(x^{4}+2 x^{3}-x^{2}-2 x\right) \equiv \\
\equiv\left[q(x)-\left(x^{2}+2 x\right)\right]\left[q(x)-\left(x^{2}-1\right)\right] .
\end{gathered}
$$

The product of non-zero polynomials cannot be the zero polynomial, so one of the two factors must be the identically zero polynomial. Therefore, $q_{1}$ and $q_{2}$ are solutions, and there are no other solutions to the problem.