For your conditions what is the most important factors will be NEE (Noise Equivalent Exposure, QE = Quantum efficiency).
The ratio \$\dfrac{SEE}{NEE}\$ is your Dynamic Range (DR) where SEE = saturation Equivalent Exposure.
You need to understand what is meant by exposure, this is the integral of the photon flux over time. In other words the # of photons collected in a time period.
Unfortunately, most sensor manufacturers Quote NEE and SEE in electrons (after the conversion of Photons to carriers - here electrons) rather than actual photons so you will need to bring in the QE to calculate the actual light levels. These numbers are often implied with a saturation level being quoted, in that case the QE is implied.
In your low light high speed application you need a sensor with as small a NEE as possible, and you will need to see in the data-sheet some mention of CDS (Correlated Double Sampling) or kTC noise removal.
After update with datasheet: *****
Using nominal Vsat with conversion gain:
\$FW = \dfrac{V_{sat}}{G_{conversion}}=\dfrac{2.7}{3.4*10^{-6}} [\dfrac{V}{\dfrac{V}{e^{-}}}] = 274,118 [e^{-}]\$ FW= Full Well
This is close enough to the 800 \$ke^{-}\$ in the data-sheet. SO the SEE = 800 \$ke^{-}\$.
Dynamic range is 71 dB which is 3548:1.
\$ NEE = \dfrac{800,000}{3548}=225 [e^-]\$
Using your dark current calculation of 1765 electrons being generated in 1 second, the noise associated with that is:
\$ \sqrt{1765} = 42 [e^-] \$
Ideally the dark current contributes a variable baseline with temperature and the noise associated with that baseline shift is the shot noise of the leakage current.
The dark shot noise and the amplifier noise being independent of each other add in quadrature:
\$ Noise_{total} = \sqrt{225^2 + 42^2} = 229[e^-]\$
Using your QE calculation from above, the NEE is \$ \dfrac{229}{0.522} = 439 \gamma\$
You can do the same with the Hamamatsu S11639.
However, you still cannot directly compare the two because you have omitted an very important datapoint. What is the area of a pixel?
What is important is that to compare these two sensors in the same conditions. You need to understand the irradiance required to meet NEE, which has units of \$\dfrac[W][m^2]\$ but \$\dfrac{\gamma}{m^2}\$ is comparable if you are using a single wavelength. Here \$\gamma\$ means photons.
Your next step in the comparison is look at the optical setup, f/#, resolution etc.
Best Answer
Sensitivity, formally, is $$\frac{\partial \text{Output}}{\partial {\Theta}} $$
where \$ \Theta\$ is the parameter of interest.
This is generally the way sensitivity appears, thought the output can be specified different ways, such as the change in resistance of a sensor, or the resulting voltage change.