You are right about the base case; 3 and 6 are given, and you've shown that the formula yields them.
But the inductive step is to assume that [imath]a_i = 2^i+i[/imath], not for every [imath]i\ge 1[/imath], which is what you are to prove, but for every [imath]1\le i\le k[/imath] for some given k, and use that to prove that it is true for the next case, namely [imath]a_{k+1} = 2^{k+1}+(k+1)[/imath].
So, what does the definition say [imath]a_{k+1}[/imath], using the formula for the two preceding terms? Try to manipulate that to look like [imath]2^{k+1}+(k+1)[/imath].
No, you don't need to show the value of [imath]a_3[/imath]; you'll be using the definition in the inductive step.