Relationship between rate of reaction and pressure
The exponential term in the Arrhenius equation does not imply reaction with other molecules. It applies to ready to react unstable adduct, about to overcome the energy barrier $E_a$ to form the product.
The concentration of this adduct is driven by concentration of reagents and therefore by the pressure.

Therefore the reaction rate equation implicitly involve 2 multiplicative terms:


Concentration of reagent adduct

$$\mathrm{const}\cdot [A]^2 = \mathrm{const}\cdot \left(\frac{p_{\rm A}}{RT}\right)^2$$
Probability the adduct reaches the reaction energy barrier, from the Boltzmann distribution:

$$\exp{\left(-\frac{E_A}{RT}\right)}$$


By other words, the constant $A$ in your equation

$$v = A \exp\left(-\frac{E_A}{RT}\right)$$

implicitly involves concentration, respectively pressure of the reagents.