how come -Q1+x charge can get induce on the second plate?(we never really gave q1/-q1 charge on second conductor )
This is answered in the community wiki answer of the linked question in the "One more preliminary" part before the actual proof. It is due to the field inside of a conductor being $0$ and then subsequent use of Gauss's law using a Gaussian pill box that has one side in the middle of each conductor.
Suppose if +Q1 and Q2 charges are given to conductors plate wouldn't all the charge go to outer surface due to electrostatic repulsion and thus no charge is left on inner side.
No, if this happened then the field within each conductor would not be $0$ in general. The community wiki answer in the linked question shows and explains how the charges would be distributed in the system.
The only way this would happen is if $Q_1=Q_2$ because, as you can calculate yourself, the charge on the inside surfaces will have a magnitude of $\frac{Q_1-Q_2}{2}$.
Shouldn't Electric field be automatically 0 as permittivity of conductors is infinite
I don't understand this point. What field are you talking about? The field within each conductor is $0$ and the field between each conductor is not $0$ due to how the charges distribute. Using the work shown in the community wiki answer you can determine what the field will be in any region you want.
