I have looked for a while for a proof which does not use the Calabi-Yau theorem and nobody seems to know it.

Also, there are plenty of non-Kaehler manifolds with canonical bundle trivial topologically and non-trivial as a holomorphic bundle (the Hopf surface is an easiest example).

The argument actually uses the Calabi-Yau theorem, Bochner's vanishing, Berger's classification of holonomy and Bogomolov's decomposition theorem.

From Calabi-Yau theorem you infer that there exists a Ricci-flat Kaehler metric. Since the Ricci curvature is a curvature of the canonical bundle, this implies that the canonical bundle admits a flat connection.

Of course, this does not mean that it is trivial holomorphically; in fact, the canonical bundle is flat on Hopf surface and on the Enriques surface, which are not Calabi-Yau.

For Calabi-Yau manifolds, however, it is known that the Albanese map is a locally trivial fibration and and has Calabi-Yau fibers with trivial first Betti number. This is shown using the Bochner's vanishing theorem which implies that all holomorphic 1-forms are parallel.

Now, by adjunction formula, you prove that the canonical bundle of the total space is trivial, if it is trivial for the base and the fiber. The base is a torus, and the fiber is a Calabi-Yau with $H^1(M)=0$. For the later, triviality of canonical bundle follows from Bogomolov's decomposition theorem, because such a Calabi-Yau manifold is a finite quotient of a product of simple Calabi-Yau manifolds and hyperkaehler manifolds having holonomy $SU(n)$ and $Sp(n)$. Bogomolov's decomposition is itself a non-trivial result, and (in this generality) I think it can be only deduced from the Berger's classification. The original proof of Bogomolov was elementary, but he assumed holomorphic triviality of a canonical bundle, which we are trying to prove.

This argument is extremely complicated; also, it is manifestly useless in non-Kaehler situation (and in many other interesting situations). I would be very interested in any attempt to simplify it.

I have looked for a while for a proof which does not use the Calabi-Yau theorem and nobody seems to know it.

Also, there are plenty of non-Kaehler manifolds with canonical bundle trivial topologically and non-trivial as a holomorphic bundle (the Hopf surface is an easiest example).

The argument actually uses the Calabi-Yau theorem, Bochner's vanishing, Berger's classification of holonomy and Bogomolov's decomposition theorem.

From Calabi-Yau theorem you infer that there exists a Ricci-flat Kaehler metric. Since the Ricci curvature is a curvature of the canonical bundle, this implies that the canonical bundle admits a flat connection.

Of course, this does not mean that it is trivial holomorphically; in fact, the canonical bundle is flat on Hopf surface and on the Enriques surface, which are not Calabi-Yau.

For Calabi-Yau manifolds, however, it is known that the Albanese map is a locally trivial fibration and and has Calabi-Yau fibers with trivial first Betti number. This is shown using the Bochner's vanishing theorem which implies that all holomorphic 1-forms are parallel.

Now, by adjunction formula, you prove that the canonical bundle of the total space is trivial, if it is trivial for the base and the fiber. The base is a torus, and the fiber is a Calabi-Yau with $H^1(M)=0$. For the later, triviality of canonical bundle follows from Bogomolov's decomposition theorem, because such a Calabi-Yau manifold is a finite quotient of a product of simple Calabi-Yau manifolds and hyperkaehler manifolds having holonomy $SU(n)$ and $Sp(n)$. Bogomolov's decomposition is itself a non-trivial result, and (in this generality) I think it can be only deduced from the Berger's classification. The original proof of Bogomolov was elementary, but he assumed holomorphic triviality of a canonical bundle, which we are trying to prove.

This argument is extremely complicated; also, it is manifestly useless in non-Kaehler situation (and in many other interesting situations). I would be very interested in any attempt to simplify it.

Update: Just as I was writing the reply, Dmitri has posted a link to Bogomolov's article, where he proves that some power of a canonical bundle is always trivial, without using the Calabi-Yau theorem.