All real coefficient polynomials $P$ satisfying the requirements of the problem are the zero polynomial and
$$P(x) = (x^2 - 2x + 3)^n \quad \forall x \in \mathbb{R},$$
where $n$ is any nonnegative integer.
Substituting into the original equation, it is easy to see that the above $P(x)$ are all solutions of this functional equation; in what follows we consider the case where $P(x)$ is not the zero polynomial.
First we prove that $P$ has no real roots. Using proof by contradiction, if $P$ has a real root $\alpha$, substituting $x = \alpha$ into the original equation gives
$$P(\alpha^2 - \alpha + 3) = P(\alpha)P(\alpha + 1) = 0.$$
Let $\beta = \alpha^2 - \alpha + 3$, then $\beta$ is also a real root of $P$, and clearly $\beta \ge 2$. Consider the infinite sequence
$$\beta, f(\beta), f(f(\beta)), f(f(f(\beta))), \dots$$
where $f(x) = x^2 - x + 3$; by the same method as above, substituting $\beta, f(\beta), f(f(\beta)), \dots$ respectively into the original equation, we obtain that every term of this sequence is a real root of $P$.
When $x \ge 2$ it is clear that $f(x) > x$, so this sequence is strictly increasing, that is, $P$ has infinitely many real roots, contradicting the fact that $P$ is a polynomial! Since $P$ has no real roots, its degree must be even. Consider the leading coefficient $t$ of $P$; comparing the coefficients of the highest degree term in the original equation gives

$t^2 = t$, so $t = 1$, that is, $P$ is a monic polynomial.
Let the degree of $P$ be $2n$, and consider $P(x) = Q(x) + (x^2 - 2x + 3)^n$; since $P$ is monic, $deg(Q) < 2n$. Substituting back into the original equation gives
$$(Q(x) + (x^2 - 2x + 3)^n)(Q(x + 1) + ((x + 1)^2 - 2(x + 1) + 3)^n) = Q(x^2 - x + 3) + ((x^2 - x + 3)^2 - 2(x^2 - x + 3) + 3)^n$$
Expanding and simplifying gives
$$\begin{aligned} & Q(x)Q(x+1) + Q(x)((x+1)^2 - 2(x+1) + 3)^n \\ & \quad + Q(x+1)(x^2 - 2x + 3)^n \\ & = Q(x^2 - x + 3). \end{aligned}$$
If $Q$ is not the zero polynomial, let $deg(Q) = q$ and let the leading coefficient of $Q$ be $k$; since $q < 2n$, the highest degree of the left side is at most $2n + q$, and the coefficient of the $2n + q$ degree term is $2k \neq 0$, so the highest degree of the left side is $2n + q$. But the highest degree of the right side is clearly $2q$, so $n = 2q$, a contradiction!
Therefore $Q$ can only be the zero polynomial, that is, all the possibilities for $P$ are only the identically zero polynomial and
$$P(x) = (x^2 - 2x + 3)^n.$$