Disclaimer: The solutions we've shared are just one exciting approach, and there are surely many other wonderful methods out there. We’d love to hear your alternative solutions in the community thread below, so let's keep the creativity flowing!

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.