Setting $y = 0$ we get
$$2P^2(x) + P^2(0) = 2P(x^2) \qquad (1)$$
for every $x \in \mathbb{R}$. We claim that $P(x)$ is a monomial. Indeed, if this is not the case, let $ax^n$ and $bx^m$, with $n > m$ be the two non-zero terms with the largest powers of $x$. Comparing the coefficients of $x^{m+n}$ in (1) we get $2ab = 0$, a contradiction.
So $P(x)$ is a monomial. If $P(x) = c$, a constant, then substituting in the original equation we get $3c^2 = 2c$ giving $c = 0$ or $c = \frac{2}{3}$.
Otherwise $P(x) = ax^n$ for some $a \in \mathbb{R} \setminus \{0\}$ and $n \in \mathbb{N}$. Then $P(0) = 0$ and substituting in (1) with $x = 1$ we get $2a^2 = 2a$ giving $a = 1$.
Now for $x = y = 1$ in the original equation we get $2 + 2^{2n} = 2 \cdot 3^n$. The cases $n = 1, 2$ are obvious solutions, while for $n \ge 3$ we have
$$\frac{2 + 2^{2n}}{3^n} > \left(\frac{4}{3}\right)^n \ge \frac{64}{27} > 2$$
showing that no other solutions exist.
So the only possible solutions are $P(x) = 0, \frac{2}{3}, x, x^2$ which are easy to check that they satisfy the equation.

If $P$ is constant, then we get $P(x) = 0$ or $P(x) = \frac{2}{3}$ as in the first solution. So assume that the degree of $P$ is $n \ge 1$ and let $a_n$ be the leading coefficient. Equating the coefficients of $x^{2n}$ we $2a_n^2 = 2a_n$ giving $a_n = 1$.
Now letting $x = y$ we get
$$2P^2(x) + P^2(2x) = 2P(3x^2)$$
and equating the coefficients of $x^{2n}$ we get $2 + 2^{2n} = 2 \cdot 3^n$ which gives $n = 1$ or $n = 2$ as in the first solution. We can now try all polynomials of the form $P(x) = x + a$ and $P(x) = x^2 + ax + b$ which leads to the same solutions as in the first solution.