Let $(a_k)_{1\leq k\leq 2023}$ be the required sequence.

\[a_k=3^{2024-k}\cdot 4^{k-1}\]

Here, the sum of the squares of two consecutive terms \[a_k^2+a_{k+1}^2=(3^{2024-k}\cdot 4^{k-1})^2+(3^{2023-k}\cdot 4^{k})^2=(3^{2023-k}\cdot 4^{k-1})^2(3^2+4^2)=(3^{2023-k}\cdot 4^{k-1}\cdot 5)^2\]

Which is a square.

So, such a sequence exists.