Yes, V_{max} is dependent on enzyme concentration from the equation you provided: V_{max} = k_{cat}[E], where k_{cat} is independent of the enzyme concentration. So, you are correct, if you double the enzyme concentration, your V_{max} 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 k_{cat}. Given that you are plotting V/[E] versus [S ], The value at which your Michaelis-Menten curve saturates will be your k_{cat} (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).

K_{m}, 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)