When n is even, $ 2^n$ is a perfect square, so we must also have n+1 a perfect square. Simple argument shows that 2k+1 and 2(k+1)+1 cannot be both perfect squares, so 2 consecutive even n cannot be in the consecutive square sequence, so there are at most 3 numbers in the sequence, Odd, Even, Odd. But then let n be odd, so let n=2k+1. Then since $ 2^{n-1}$ and 2^{(n+2)-1} are perfect square, we must have 2(n+1) and 2((n+2)+1) both perfect squares greater than 0. This again cannot happen by a simple argument, so we can have at most 2 perfect squares in the sequence. This can happen, let n=48,49 for example.