Yes, Vmax is dependent on enzyme concentration from the equation you provided: Vmax = kcat[E], where kcat is independent of the enzyme concentration. So, you are correct, if you double the enzyme concentration, your Vmax should double.
Since the rates you report have already been divided by the amount of enzyme, you do not need to divide by the enzyme concentration again to get kcat. Given that you are plotting V/[E] versus [S ], The value at which your Michaelis-Menten curve saturates will be your kcat (note that the moles of product and moles of enzyme can be thought to cancel assuming one active site per enzyme, giving units of inverse time).
Km, however, is not dependent on the concentration of enzyme.
(edit: renge ishyo beat me to, but I'll leave my explanation anyway in case it helps)