Let $f(2a^k) = g(a)f(a) + h(a)$ where the degree of $h$ is less than the degree of $f$, or $h$ is the zero polynomial. Thus $f(a) | h(a)$ for all positive integers $a$ such that $f(a) \neq 0$. If $h$ is not the zero polynomial, the degree of $h$ is less than the degree of $f$, so for sufficiently large $a$, $f(a), h(a) \neq 0$ and $|f(a)| > |h(a)|$ so $f(a)|h(a)$ does not hold. Contradiction! Thus $h$ is the zero polynomial, so $f(2a^k) = g(a)f(a)$. (this is an identity) Thus if $\alpha$ is a root of $f$, $2\alpha ^ k$ is also a root of $f$. Now define the sequence $x_1 = \alpha, x_{m+1} = 2(x_m)^k$ for all positive integers $m$. Clearly, $x_1, x_2, \cdots$ are all roots of $f$. Consider $|\alpha|$. Case 1: $|\alpha| = 0$. Then $\alpha = 0$. Case 2: $|\alpha| > 0.5^{\frac{1}{k-1}}$. Then $|x_1| < |x_2| < \cdots$ so $x_i$ are pairwise distinct so $f$ has infinitely many roots so it is the zero polynomial, contradiction! Case 3: $|\alpha| < 0.5^{\frac{1}{k-1}}$. Then $|x_1| > |x_2| > \cdots$ so $x_i$ are pairwise distinct so $f$ has infinitely many roots so it is the zero polynomial. Contradiction again! Case 4: $|\alpha| = 0.5^{\frac{1}{k-1}}$. Potentially possible. Summing up, $\alpha = 0$ or $|\alpha| = 0.5^{\frac{1}{k-1}}$. If $f$ has at least one root $\alpha$ such that $|\alpha| = 0.5^{\frac{1}{k-1}}$, the modulus of the coefficient of the power, that is the lowest power with a nonzero coefficient, must be a positive integer multiple of $0.5^{\frac{1}{k-1}}$. However, this can never be an integer, contradiction! Thus all roots of $f$ are $0$ so $f = a^n$ is the only possibility.

Isn't $f(x)=cx^n$ possible too?

Nice and easy. Let us consider $f(2X^k)=g(X)$ as a polynomial in variable $X$. By the division lemma of Euclid, we can write $g(X)=f(X)\cdot q(X)+r(X)$ with degree of $r$ less than degree of $f$, where $q,r$ are polynomials with rational coefficients. Since $f(a) \mid f(2a^k)$ for all $a$ sufficiently large, we see that multiplying by the common denominators of $q,r$, say, $N \not= 0$, we get that $f(a) \mid Nr(a)$ for all $a$ sufficiently large. However, this fails to hold as $\frac{Nr(a)}{f(a)} \rightarrow 0$ as $a \rightarrow \infty$, yielding that $r=0$. Thus, $f(X) \mid Nf(2X^k)$ (in the ring $Z[X]$!). Let $b$ be the largest non negative integer such that $X^b \mid f(X)$. Let $f(X)=X^bh(X)$ and see that $h(X) \mid 2^kNh(2X^k)$; let $\alpha$ be a root of $h$, then so is $2\alpha^k$. Hence, if $|\alpha| \ge 1$ then $h$ has infinitely many roots, a contradiction! Since all roots of $h$ are of modulus less than $1$ and $h$ has integer coefficients, we have $h(0)=0$, a contradiction! Thus, $h$ must have a zero degree (and hence, no roots!) and we conclude that $f(X)=cX^b$ where $c,b$ are integers, $b \ge 0$. These clearly satisfy our hypothesis.