Resonant Q Factor vs Voltage Gain
If I have a resonant system, How does Q affect peak resonant voltage,
i.e. does a doubling of Q result in double the power being stored in
the system?

If you are considering applying a sine wave to a highly resonant tuned circuit then, at resonance, the output voltage is the input voltage multiplied by \$Q\$. Consider this RLC circuit: -

As for the math, it's fairly straightforward to show that a low pass RLC filter (for example) has a transfer function equation thus: -
$$H(j\omega) = \dfrac{1}{1-\frac{\omega^2}{\omega_n^2}+j\frac{2\zeta\omega}{\omega_n}}$$
And, at resonance \$\omega = \omega_n\$ hence: -
$$H(j\omega) = \dfrac{1}{1-\frac{1}{1}+j\frac{2\zeta}{1}} = \dfrac{-j}{2\zeta} = -jQ$$
So, at resonance the transfer function has a magnitude of \$Q\$ phase shifted by 90°. See the orange circles below: -

And, given that we are only interested in highly resonant circuits (where Q is quite or very large), we can say without fear of significant error, that an accurate measure of the peak of resonance has an amplitude (\$V_{OUT}\$) that is \$Q \times V_{IN}\$.
However, if you take power from \$V_{OUT}\$ then you are taking current and thus the &quot;loading&quot; acts like a resistor. This means that Q is then subject to the following formula: -
$$Q = R_{load}\cdot\sqrt{\dfrac{C}{L}}$$
And, in all these systems, power-out equals power-in minus losses. There is no power gain; only voltage gain.
So, in summary, \$Q\$ defines the output voltage amplification at resonance and \$Q\$ is determined by the capacitance, inductance and load resistance.