Since $n+1$ values already uniquely determine a polynomial of degree $n$, and
$$f(k) = 2^k = (1+1)^k = C(k, 0) + C(k, 1) + \dots + C(k, n)$$
holds for $k = 0, 1, \dots, n$, while the right-hand side is a polynomial of degree $n$, we know that
$$f(x) = C(x, 0) + C(x, 1) + \dots + C(x, n).$$
Moreover, since $n$ is odd, pairing up terms two by two in the above expression, we obtain
$$\begin{aligned} f(x) &= C(x + 1, 1) + C(x + 1, 3) + \dots + C(x + 1, n) \\ &= (x + 1) \left[ 1 + \frac{1}{3}C(x, 2) + \frac{1}{5}C(x, 4) + \dots + \frac{1}{n}C(x, n - 1) \right]. \end{aligned}$$
Let $n!f(x) = (x+1)R(x)$; note that $R(x)$ is a polynomial with integer coefficients. For all integers $x$, we have
$$\text{gcd}(x + 1, R(x)) \bigg|_{R(-1)} = n! \left[ 1 + \frac{1}{3} + \frac{1}{5} + \dots + \frac{1}{n} \right].$$
Note that $R(-1)$ is a nonzero integer, so $\nu_2(R(-1))$ must be finite, where
$$\nu_2(m) := \sup \{k : 2^k | m\}.$$
In other words, we have
$$\begin{aligned} & \min\{\nu_2(x+1), \nu_2(R(x))\} \\ &= \nu_2(\text{gcd}(x+1, R(x))) \le \nu_2(R(-1)) < \infty. \end{aligned} \quad (1)$$

Now, suppose $x$ is an integer such that $f(x)$ is a power of 2, then we have
$x + 1|n! \times 2^{\nu_2(f(x))}$ and $R(x)|n! \times 2^{\nu_2(f(x))}$. But by Eq. (1), this means that
$$x + 1 \left| n! \times 2^{\nu_2(x+1)} \right| n! \times 2^{\nu_2(R(-1))} \qquad (2)$$
and
$$R(x) \left| n! \times 2^{\nu_2(R(x))} \right| n! \times 2^{\nu_2(R(-1))} \qquad (3)$$
at least one must hold. However, since
$$\lim_{|x| \to \infty} |x + 1| = \infty \quad \text{and} \quad \lim_{|x| \to \infty} |R(x)| = \infty,$$
it is easy to see that the $x$ satisfying at least one of Eq. (2) and Eq. (3) can be at most finitely many. This completes the proof.