How do we know sound pressure squared is proportional to the power in the sound wave? I am trying to find the equation that explains why we can say the pressure exerted by a wave (squared) is proportional to the power. I see this claim made a lot but I can't tell where it comes from.
 A: First, let's consider a pressure transducer, such as a microphone.
When we use it to record sound, in practice we are logging a voltage. For a linear system, each increase in pressure results in an equal increase in voltage. For electrical power, $P =  V^2/R$.
A convention is to ignore the resistance/impedance (i.e., $R$, in Ohms) by setting $R = 1$. This leads to reporting the measured value as proportional to the actual acoustic power (see Norton and Karczub 2003, Fundamentals of Noise and Vibration Analysis for Engineers):

...the spectra of continuous signals are referred to as power spectral
densities because they have units related to power and that the
spectra of transient signals are referred to as energy spectral
densities because they have units relating to energy. This is an
important point - one which warrants further discussion. A power
spectral density has units of (volts)^2 per hertz or V^2 s. Thus the
area under a power spectral density curve has units of (volts)^2 which
is proportional to power (i.e. electrical power is ∝ V^2).


Now we proceed by analogy to actually answer your question.
Matter resists acoustic flow, a physical quantity we call acoustic impedance, $Z$. Impedance varies depending on material properties (such as mass, elasticity, and geometric configuration.) So we can write the above equation for a purely acoustic scenario as:
$$P = \frac{p^2}{Z}$$
Or, accepting that $Z$ is variable (and often empirically determined), it may be simpler to say $P \propto p^2$.
