Only $P\equiv 0$ satisfies. Suppose that $P$ is nonconstant. We have $rad(n^2+2^n)|P(n^2)$. WLOG we can assume that the leading coefficient is positive. The polynomial is not constant hence we can pick $m$ with $P(m^2)>0$. Consider the sequence $a_l=2^l$ and $b_l=a_l+m^2$. Since the sequence $\{a_i\}$ has one prime divisor, by Kobayashi we conclude that there are infinitely many primes dividing at least one term of $\{b_i\}$. Choose $q>P(m^2)$ and $q|m^2+2^l$. Also take $k\equiv l-m(mod \ q-1)$ which implies $q|m^2+2^{m+k}$ or $q|m^2+2^{m+qk}$ or $q|(m+qk)^2+2^{m+qk}$. This yields $q|P((m+qk)^2)\iff q|P(m^2)$. However this is impossible because of $q>P(m^2)>0$ which results in a contradiction.$\blacksquare$

Let us observe that \( \lfloor\sqrt{n}\rfloor = k \) for any two positive integers \( n \) and \( k \) such that \( n \in [k^2, k^2 + 2k] \). Fix a positive integer \( m \geq N \) and a prime number \( p \). Set \( k = p - 1 + m \) in the above and consider the interval \( [k^2, k^2 + 2k] \). This interval has a length of \( 2k > p \), so it contains a positive integer \( n \equiv -2^m \pmod{p} \). Then, by Fermat's Little Theorem, we have: \[ n + 2^{\lfloor\sqrt{n}\rfloor} = n + 2^k \equiv -2^m + 2^{p-1+m} \equiv 0 \pmod{p}. \]Therefore, \( p \mid P(n) \). Since \( P \) is a polynomial with integer coefficients, we deduce: \[ P(-2^m) \equiv P(n) \equiv 0 \pmod{p}. \]Thus, \( p \mid P(-2^m) \) for every prime number \( p \). Consequently, \( P(-2^m) = 0 \) for every positive integer \( m \geq N \). This implies that \( P \equiv 0 \), completing the proof.