It is easy to see that all the $x_i$ $i>1$ take $3$ values ${0,2x_1,-2x_1}$ by pigeonhole $k=675$ and construction is trivial

Rather nice and easy algebra problem. We start off by characterizing all such sequences. Claim : The terms $x_i \in \{0,2x_1,-2x_1\}$ for all $2\leq i\leq 2024$. Proof : Let $S_i =x_1+\dots + x_i$ for all $1\leq i \leq 2024$. Then, we show that $S_i \in \{x_1,-x_1\}$ for all $1\leq i \leq 2024$. For this first note that the given condition is, \begin{align*} (x_1+\dots + x_i)^2 &= (x_1+x_2+\dots + x_{i+1})^2\\ (x_1+\dots + x_i)^2 &= (x_1+\dots + x_i)^2 + 2x_{i+1}(x_1+\dots + x_{i}) + x_{i+1}^2\\ x_{i+1}^2 &= -2x_{i+1}(x_1+\dots + x_{i}) \end{align*}Thus, for all $1 \leq i \leq 2023$, $x_{i+1}=0$ or $x_{i+1}=-2(x_1+\dots +x_i)=-2S_i$. Then, we proceed via induction. It is clear that $S_1=x_1$. Say the claim is true for some $k\geq 1$. Then, $S_{k+1}=S_{k}+x_{k+1}$ If $x_{k+1}=0$, we immediately have $S_{k+1}=S_k \in \{x_1,-x_1\}$. Else, $x_{k+1}=-2S_k$ by our previous observation. Thus, $S_{k+1}=-S_k \in \{x_1,-x_1\}$ and thus we are done via induction. Now, if $x_i \neq 0$ for some $2\leq i \leq 2024$ , we know that $x_i = -2S_{i-1}$. Since $S_{i-1} \in \{x_1,-x_1\}$, it follows that $x_i \in \{-2x_1,2x_1\}$ which proves the claim. Now, if $x_1 \neq 0$, all 3 of $0,2x_1,-2x_1$ are distinct. Thus, excluding the first term and by Pigeon hole, it follows that there are atleast, \[\lceil \frac{2024-1}{3} \rceil = 675\]equal terms. To see that we can't always do any better, simply consider the sequence, $x_1=1$ and \[x_i = \begin{cases} 0 & \text{ if } i \equiv 2 \pmod{3}\\ -2 & \text{ if } i \equiv 0 \pmod{3}\\ 2 & \text{ if }i \equiv 1 \pmod{3} \end{cases} \]for all $2 \leq i \leq 2024$.