I asked a colleague who works in complex analysis and he also does not know any "integration free" proof or argument. However, he pointed me to a version free of complex numbers but equivalently astonishing (at least for me): Weyl's Lemma, stating that any function $u\in L^1_\text{loc}$ satisfying $$\int u\Delta \phi = 0$$ for any test function $\phi$ is $C^\infty$. However, the proof also uses integration in the form of convolutions and hence, integrals are not at all avoided.
I asked a colleague who works in complex analysis and he also does not know any "integration free" proof or argument. However, he pointed me to a version free of complex numbers but equivalently astonishing (at least for me): Weyl's Lemma, stating that any function $u\in L^1_\text{loc}$ satisfying $$\int u\Delta \phi = 0$$ for any test function $\phi$ is $C^\infty$. However, the proof also uses integration in the form of convolutions and hence, integrals are not at all avoided.