Obviously $n=1$ doesn't work. So suppose $n>1$. Case 1: $n$ is even. Since $2^n$ and $2^{n+2}$ are both perfect squares, $n+1$ and $n+3$ are both perfect squares. But one of them must be $\{2,3\}\pmod4$, which is a contradiction. Case 2: $n$ is odd. Then $2(n+1)$ and $2(n+3)$ are both perfect squares. But one of $\{n+1,n+3\}$ is $2\pmod4$, which implies one of $\{2(n+1), 2(n+3)\}$ is $4\pmod8$, and thus can't be a perfect square.

Assume that there is a number n that satisfies the above, then the product of the two has to also be a perfect square. So, $(n+1)(n+3)2^{2(n+2)}$ has to be a perfect square. It suffices to show $(n+1)(n+3)$ is a perfect square Taking to consideration that: $(n+2)^2\geq (n+1)(n+3)\geq(n+1)^2$ Its impossible to have $=$ in any case because $n\geq1$

For the sake of contradiction, suppose that there exists a positive integer $n$ such that $2^n(n+1)$ and $2^{n+2}(n+3)$ are both perfect squares. This implies that their product $$\left(2^{n+1} \right)^2 (n+1)(n+3)$$is a perfect square. In order for this to be a perfect square, $(n+1)(n+3)=n^2+4n+3$ must be a square. However, we notice that $$(n+1)^2 < (n+1)(n+3) < (n+2)^2,$$implying that $(n+1)(n+3)$ is not a perfect square. This is a contradiction.

Just note that (n+3)(n+1)=(n+2)^2-1

If they were both perfect squares, their product is a perfect square. Then 2^(2n+2)*(n+3)(n+1) is a perfect square, 2n+2 is even so perfect square needs to be (n+3)(n+1)=(n+2)^2-1, contradiction, since there are no two squares differing by one that satisfy the problem.

Assume, $n>1.$ Case 1: $n,$ is even. Then, we have $2^n, 2^{n+2}$, are both perfect squares, but $n+1, n+3$, can not be perfect squares because $n+3-(n+1)=2,$ and the minimum, difference, can be $2^2-1^2=3,$ hence, contradiction. Case 2: $n,$ is odd. Then, $2n+2, 2n+6,$ need to be perfect squares. Note, that the difference, between perfect squares, is $4.$ Now, $2^2-1^2=3, 3^2-2^2=5,$ and the differences, will keep on getting bigger than $5,$ hence, contradiction.