Let $f(z)$ be an entire function of finite order of the form $$f(z)=\sum_{n=0}^\infty a_nz^{k_n}$$ where $a_n\neq0$ for all $n$ and $(k_n)_{n\geq0}$ is an increasing sequence of integers satisfying $$\limsup_n\;(k_{n+1}-k_n)=\infty.$$ Show that $f(z)$ has infinitely many zeros.
I know that if $f(z)$ has only finitely many zeros then by Hadamard's theorem $f(z)$ is of the form $P(z)e^{Q(z)}$ for some polynomials $P$ and $Q$, so a possible approach is to show that for such functions the condition $\limsup_n\;(k_{n+1}-k_n)=\infty$ is not possible, but I gave up after some time. Another possible way may be trying to prove that $f(z)$ has non-integer order, since in such a case it follows easily that $f(z)$ cannot have finitely many zeros. Any ideas?
As you said, if $f$ has only finitely many zeros, it is of the form $f(z) = P(z) e^{Q(z)}$ with polynomials $P(z) = p_m z^m + \ldots + p_0$ and $Q(z) = q_n z^n + \ldots + q_0$. (All leading coefficients are assumed to be non-zero.) Then $P(z) f'(z) = R(z) f(z)$ with $R(z) = P'(z) + P(z)Q'(z) = r_N z^N + \ldots + r_0$, where $N=m+n-1 \ge m$. Writing this out in coefficients, you get $$ (p_m z^m + \ldots + p_0) \sum_{j=0}^\infty k_j a_j z^{k_j-1} = (r_N z^N + \ldots + r_0) \sum_{j=0}^\infty a_j z^{k_j} = \sum_{k=0}^\infty b_k z^k $$ Now pick some $j$ such that $k_{j+1} - k_j > N+2$ and consider the largest $k \in \{ k_j, k_j +1 , \ldots, k_{j+1}-2\}$ such that $b_k \ne 0$. Multiplying out terms we have $k = m+k_j-1 = N+k_j$, which implies that $N = m-1$, contradicting $N \ge m$.