As $P(x) = 1$ has no real root, $x^{f(0)}$ has also no real root. So $f(0) = 0$.
Now considering $x^i + 1$, it has a real root if and only if $i$ is odd. So $i$ and $f(i)$ have the same parity.
For even $n$ consider the following polynomial in which $a, \epsilon$ are positive numbers:
$$P(x) = x^n + n a^{n-1} x + (n-1) a^n + \epsilon$$
This polynomial has no real root, indeed applying AM-GM inequality yielding
$$x^n + a^n + \cdots + a^n \ge n x a^{n-1} |x|.$$
Then by the problem's condition, the polynomial
$$Q(x) = x^{f(n)} + n a^{n-1} x^{f(1)} + (n-1) a^n + \epsilon$$
has no real root. Thus $q(-a^{\frac{1}{f(1)}})$ should be positive yielding:
$$a^{\frac{f(n)}{f(1)}} + \epsilon > a^n \implies a^{\frac{f(n)}{f(1)}} \ge a^n$$
By putting $a$ greater and less than 1 we get $f(n) = n f(1)$.
For odd $n$ consider the polynomial $(x^n + a)^2 + \epsilon$ for a positive $a$ and $\epsilon$. This polynomial is obviously positive so
$$x^{2n f(1)} + 2a x^{f(n)} + a^2 + \epsilon > 0$$
By putting $x = -a^{\frac{1}{f(n)}}$ we have:
$$a^{\frac{2n f(1)}{f(n)}} + a^2 + \epsilon > 2a^2 \implies a^{\frac{2n f(1)}{f(n)}} \ge a^2 \implies f(n) = n f(1).$$
This finishes our proof.