This seems to be an old remark by Andreotti and appears as exercise 2.6.10 in the book `Complex Geometry`

by D. Huybrechts. The tangent bundle of $M$ is complexified
and $\omega$ is viewed as a complex $2$-form. An almost complex structure is equivalent
to a decomposition $TM^{\mathbf C}=T^{1,0}M\oplus T^{0,1}M$, where $T^{0,1}M=\overline{T^{1,0}M}$. If $\omega$ is to be holomorphic, then it must vanish on $T^{0,1}M$. This will give uniqueness.

The conditions $\omega\wedge\omega=0$ and $\omega\wedge\bar\omega$ a volume form
imply that the kernel of $\omega$ is $2$-dimensional, so can define $\ker\omega=T^{0,1}M$, and this will give existence. In fact, by skew-symmetry the rank of $\omega$ is even; the first condition implies that $\omega$ is not symplectic so the kernel is non-trivial; the second condition implies that it cannot be $4$-dimensional, hence the claim.

Next we let $T^{1,0}M$ be the complex conjugate of $T^{0,1}M$, that is, $\ker\bar\omega$.
The condition $\omega\wedge\bar\omega$ never zero says that $T^{1,0}M\cap T^{0,1}M=0$,
so we have the desired decomposition.

Finally, to check the integrability, it suffices to see that $T^{0,1}M$ is involutive.
Let $X$, $Y$ be sections of $T^{0,1}M$ and $Z$ be an arbitrary section of $TM^{\mathbf C}$. Recall the formula
$d\omega(X,Y,Z) = X\omega(Y,Z)-Y\omega(X,Z)+Z\omega(X,Y) - \omega([X,Y],Z)+\omega([X,Z],Y)-\omega([Y,Z],X) $.

We have $d\omega=0$ and all the terms on the RHS vanish by our choice of $X$, $Y\in\ker\omega$ except $d\omega([X,Y],Z)$. Hence this term also vanish and this means that
$[X,Y]$ lies in $\ker\omega$.