Suppose such polynomials exist and write $Q(x)=a x^{2}+b x+c$. If we plug in 2, 3, 5, and 7 into $Q$, exactly the (at most) two roots of $P$ must come out. Since there cannot be more than two identical values from $Q(x)$ (because $Q$ is quadratic), we thus get two different values, each exactly twice.
Now suppose $Q(n)=Q(m)$ for different numbers $m$ and $n$. Then $a n^{2}+b n+c = a m^{2}+b m+c$, so $a\left(n^{2}-m^{2}\right)=b(m-n)$, thus $a(n+m)(n-m)=-b(n-m)$. Since $m-n \neq 0$, this implies $a(n+m)=-b$, or $n+m=\frac{-b}{a}$.
We know that we can find two different pairs $(m, n)$ and $(k, l)$ from 2, 3, 5, and 7 such that $Q(m)=Q(n)$ and $Q(k)=Q(l)$. Therefore, $m+n=\frac{-b}{a}=k+l$. We must thus be able to divide the four numbers 2, 3, 5, and 7 into two pairs that have the same sum. This is impossible, however, since $2+3+5+7=17$ is odd. We conclude that there are no polynomials $P$ and $Q$ with the desired properties.