So that must imply that with increased vapor pressure, the mole fraction of the gas in the solution must increase, but quite the opposite is true.
Vapor pressure, or saturation pressure, $P_{sat}$, is different than system pressure, $P_{sys}$. $P_{sat}$ is an intensive property, relative to the compound of interest, while $P_{sys}$ is an extensive property related to the system in which compound resides.
For Henry's Law, $$c_i = H(T)p_i$$ the pressure of interest is not the vapor pressure, but the system pressure. The partial pressure in Henry's law, $p_i$, is related to the system pressure, $p_i=y_iP_{sys}$.
Your intuition about increased vapor pressure leading to a decrease in the solubility of a component in solution is correct. Generally, if $P_{sat}(T) > P_{sys}$, then the component of interest will reside in the gas phase for a two-phase system (liquid & gas). Generally however, for gases like $\ce{N_2}$, $\ce{O_2}$, $\ce{CO_2}$, etc. the vapor pressure of the compound is already sufficiently high to begin with, i.e $P_{sat}(T) >> P_{atm}$. For example, the vapor pressure of $\ce{CO_2}$ is near $60$ atm at $25$ degC. This is where the role of the system pressure, $P_{sys}$, plays an important role: as you increase $P_{sys}$ sufficiently, you lessen the gap between $P_{sys}$ and $P_{sat}(T)$; this meaning that you will increase the prevalence of a component in solution.
Is it the case that the growth the Henry's constant outgrows the increase in partial pressure of the gas, so that the mole fraction of the gas in the solution decreases?
The Henry's law coefficient's relationship for temperature (for a limited temperature range) can be described by the following Van't Hoff relationship,
$$H(T) = H_oexp[\frac{-\Delta_{sol}H}{R} (\frac{1}{T}-\frac{1}{T_R})]$$
Where, $\Delta_{sol}H$, is the Enthalphy of dissolution (generally a negative term), $H_o$, is the Henry's constant at STP, and $T_R$, is the reference temperature (usually 298.25 degC). With this, we can re-write Henry's equation as,
$$c_i = H(T)p_i = p_iH_oexp[\frac{-\Delta_{sol}H}{R} (\frac{1}{T}-\frac{1}{T_R})] = y_iP_{sys}H_oexp[\frac{-\Delta_{sol}H}{R} (\frac{1}{T}-\frac{1}{T_R})] $$
We can see here that it is not that the Henry's constant outgrows the partial pressure; it is actually quite the opposite. The effect of increasing temperature actually lowers the value of $H(T)$, thus lowering $c_i$.
