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.

Another comment: Many theorems about (unsespected) smoothness of solutions of partial differential equation use some integral formula to deduce higher smoothness. One exception is the Cauchy–Kowalevski theorem but I don't see how this is related here.

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.

Another comment: Many theorems about (unsespected) smoothness of solutions of partial differential equation use some integral formula to deduce higher smoothness. One exception is the Cauchy–Kowalevski theorem but I don't see how this is related here.