Let $\Phi(x)$ be a free scalar field. If $\Phi^2(x)$ were a well-defined operator then the commutator $[\Phi(x),\Phi(y)]$ would vanish in the limit $y\to x$, while in fact it diverges. The latter follows by considering the Fourier transform; see, e.g., the beginning of the QFT book by Peskin/Schroeder.