You can use OpenArgs, something like this:

In Form1, when calling Form2

DoCmd.OpenForm "Form2", , , , , , "Enable"

In Form2

Private Sub Form_Load()
If Len(Nz(Me.OpenArgs, "")) > 0 Then
 If Me.OpenArgs = "Enable" Then
   YourCommandButton.Enable = True
 End If
End If
End Sub

I would use 'Not Visible' because I don't think it to be a good user
experience to have a button that just simply does not work....or an
alternative would be to trigger a message box explaining why it is not
working....

Since opening from the Main Menu may occur first...  I would set the
button's property to Not Visible.  Therefore they open Form-2 direct from
main menu and it will never be Visible....

Thinking about Form-1 ....probably there is a field in this form that has
always a value greater than 0, perhaps the key index...

So that in Form-2 in the 'Open' event you can put in
If Forms!Form-1.ControlName >1 then
Me.Button.Visible = True
End If

So in this case Form-1 is open (and therefore presumably Form-2 is being
opened by Form-1).  Of course this presumes knowing behavior of user and that
Form-1 would not be opened when they go to Form-2 from the Main Menu....but
overall you probably get the idea and can modify this to be better suited for
you.....