Let

$P\left(x\right)=a_n{x}^n+a_{n-1}{x}^{n-1}+a_{n-2}{x}^{n-2}+\dots$

$+a_{1}x+a_0$

$a_0,a_1,a_2\dots \in I$

Given, $P\left(2\right)=2$ and $P\left(4\right)=5$

$2=a_n{2}^n+a_{n-1}{2}^{n-1}+a_{n-2}{2}^{n-2}+\dots$

$+a_{1}2+a_0\dots$ (i)

$5=a_n{4}^n+a_{n-1}{4}^{n-1}+a_{n-2}{4}^{n-2}+\dots$

$+4a_1+a_0\dots$ (ii)

On subtracting Eq. (i) from Eq. (ii), we get

$3=a_n\left(4^n-2^n\right)+a_{n-1}\left(4^{n-1}-2^{n-1}\right)$

$+\dots +2a_1$

Clearly, $LHS$ is odd number and $RHS$ is even number.

$\therefore$ No polynomials exists.