Suppose $P(x) = a_0x^n + a_1x^{n-1}+\cdots+a_{n-1}x+a_n$ is such a polynomial with degree $n\ge 1$. Then $$a_0(2x^2-1)^n+a_1(2x^2-1)^{n-1}+\cdots+a_{n-1}(2x^2-1)+a_0=\frac{1}{2}\times(a_0x^n + a_1x^{n-1}+\cdots+a_{n-1}x+a_n)^2$$ Comparing the coefficient of $x^{2n}$ leads to conclusion that $a_0\times 2^n = \frac{1}{2}\times a^2_0$. Therefore $a_0=2^{n+1}$ which is a rational number. Now we are going to show all coefficients are rational. If this is not the case, let's assume that $a_{k+1}$ is the first irrational number in the sequence of $\{a_0, a_1, \cdots, a_n\}$ where $0 \le k \le n-1$. Comparing the coefficients of $x^{2n-k-1}$, the left side yields a rational number involving $a_0, a_, \cdots, a_k$ and the right side yields a number of $a_0a_{k+1}$. Therefore $a_{k+1}$ must be rational. In another word, all coefficients must be rational. However, letting $x=1$ leads to $P(2\times 1^2-1)=\frac{(P(1))^2}{2}\implies P(1)=1\pm\sqrt{3}=a_0+a_1+\cdots + a_n$ which is a contradiction. Therefore $P(x)$ must be a constant.