How to create planar acoustic waves? The only way I can think of for creating planar acoustic waves is using a 2D phased array of transmitters, but even then the planar wave is not so planar (depends on the interference pattern, which depends on the spatial resolution/pitch/size of the transmitters).
How can I actually create planar acoustic waves?
 A: Create a spherical wave really far away from your measuring device.
Since the greens function for the acoustic wave equation in a homogenous medium is a spherical wave, any acoustic source can be represented a superposition of spherical waves.
A plane wave can't be represented by a finite superposition of spherical waves, so you would need an infinite plane source to create your true plane wave. 
The realistic option is just to create a spherical wave far away from your measurement system or measure the wave very close to a finite plane source. 
This is actually somewhat practical in underwater acoustics, where long range propagation is quite common. At 1000 km from the source, the wave front is a circle of radius 1000 km. Even if my hydrophone array (underwater microphones) spans the entire water column, which is 4 km, the difference between a spherical wave and plane wave is not significant.
Considering just a cross section in the plane of my array, the wavefront of the spherical wave generated by a point source at 2km depth is represented by a circle: $$ r(z) = 1000 \text{km} \sqrt{1 - ((z-2)/1000)^{2}} $$
where z is in kilometers.  At the surface, the wavefront differs from its value at the 2km depth by a factor of  $ \sqrt{1 - (.002)^{2}} = 0.999998 $.
So it's a plane wave to good approximation. At 10 km it's a factor of $ .98 $, which is still good enough for most purposes and possible closer than the uncertainty in your hydrophone positions.   
The important non-dimensional number here is $ z/R$, where z is the span of your measuring system and R is the distance from the source (note, this is just a point source, not a fancy square speaker).  
A: A real plane wave needs to be infinitely wide, due to Fraunhofer diffraction.  We can understand this in terms of the wave being a superposition of a continuum of spherical waves, per Huygens principle. Propagation a wave via Huygens principle including from the edges of an aperture shows that the wave becomes definitely non-planar after it has traveled some distance.
So, at first it seems impossible to generate a plane wave without an infinitely wide speaker.
However, it is possible to generate a wave that very nearly approximates a plane wave if we set up a system that acts as if it is infinitely wide.  This is what a waveguide does.  A way to visualize it is as a rectangular box with mirrored sides, square in cross section (x and y) but very long (z).  Think about a light wave instead of a sound wave. At one end of the box, imagine a phased array of light sources (e.g., a collimated laser beam entering that end of the box through an array of pinholes).  The mirrored sides of the box form reflected images of the pinhole array stretching out to infinity in the x and y directions, forming what is effectively an infinitely wide plane of pinhole sources.  The wave that propagates, then, in the Z direction is the result of all those point sources.  
Change the pinhole array so that the pinholes are extremely close together, and the basics do not change.  In the limit, the pinholes can be so close together that we can dispense with them and just use the original collimated beam: the basics do not change and there is none of the diffraction that would result from the finite size & spacing of pinholes; and the result is a very nearly perfect plane wave.
Sound waves will act basically the same as light waves; we just need to make the walls reflect sound waves with very high efficiency.
When this is done with microwaves or light it's called a waveguide.  There is a lot more to be said about waveguides, and there are important differences between acoustic waves and light waves (or microwaves) that affect the forms of waves that can propagate in the waveguides, but probably that's beyond the scope of your question.
