Similar to Korea 2022: https://artofproblemsolving.com/community/c6h2830171p25041488

There's the solution in the following website: https://drive.google.com/drive/folders/1q72AafnF6OLP71U2c0YcD1OTv1JHeB9s

The answer is all $m\in\mathbb{Z_+^*}$ such that $\nu_{11}(m)\leq1$. The main idea is that we can reduce the problem to prime powers by CRT, find solutions for each prime and use Hensel's Lemma to lift the solutions for higher powers fixing one of the variables. We prove a bunch of lemmas. $\textbf{Lemma 1:}$ If a prime $p\nmid 2024$ and $p\ne5$, $\exists(x_0,y_0)\in\mathbb{F}_p^2$ such that $p\mid x_0^2+5y_0^2+2024$ and $(x_0,y_0)\ne(0,0)$. $\textit{Proof.}$ We want to find $x_0$ and $y_0$ such that $x_0^2\equiv-5y_0^2-2024(\bmod\; p)$. Since $p\notin \begin{Bmatrix}2, 5, 11, 23\end{Bmatrix}$, the map $y_0^2\rightarrow-5y_0^2-2024$ is bijective in $\mathbb{F}_p$ and covers exactly $\frac{p+1}{2}$ residues, and so does $x_0^2$. Then, by the Pigeonhole Principle, there are some $x_0,y_0$ such that $x_0^2\equiv-5y_0^2-2024(\bmod\; p)$, as we wanted. Notice that $(x_0,y_0)=(0,0)$ implies $p\mid 2024$, but we have already excluded this case. $\square$ $\textbf{Lemma 2:}$ If a prime $p\nmid 2024$ and $p\ne5$, then $\forall k\in\mathbb{Z_+}\exists(x_k,y_k)\in\mathbb{F}_{p^k}^2$ such that $p^{k+1}\mid x_k^2+5y_k^2+2024$. $\textit{Proof.}$ We know $(x_0,y_0)\ne(0,0)$. If $x_0\ne0$, consider the polynomial $P(x)=x^2+5y_0^2+2024$. Since $P(x_0)\equiv0(\bmod\; p)$ and $P'(x_0)=2x_0\not\equiv0(\bmod\; p)$, Hensel's Lemma guarantees the existence of $x_k$ for every $k$ and $y_k=y_0$ is already fixed. f $y_0\ne0$, fix $x_0$ and consider the polynomial $P(y)=x_0^2+5y^2+2024$. Since $P(y_0)\equiv0(\bmod\; p)$ and $P'(y_0)=10x_0\not\equiv0(\bmod\; p)$, we can apply Hensel again and finish. $\square$ $\textbf{Lemma 3:}$ $\forall k\in\mathbb{Z_+}\exists(x_k,y_k)\in\mathbb{F}_{5^k}^2$ such that $5^{k+1}\mid x_k^2+5y_k^2+2024$. $\textit{Proof.}$ Notice that $(x_0,y_0)=(1,0)$ is a solution $(\bmod\; 5)$. Fix $y_k=0$ and consider the polynomial $P(x)=x^2+2024$. Since $P(1)\equiv0(\bmod\; 5)$ and $P'(1)=2\not\equiv0(\bmod\; 5)$, we can just apply Hensel again. $\square$ $\textbf{Lemma 4:}$ $\forall k\in\mathbb{Z_+}\exists(x_k,y_k)\in\mathbb{F}_{23^k}^2$ such that $23^{k+1}\mid x_k^2+5y_k^2+2024$. $\textit{Proof.}$ Notice that $(x_0,y_0)=(7,2)$ is a solution $(\bmod\; 23)$. Fix $y_k=2$ and consider the polynomial $P(x)=x^2+2044$. Since $P(7)\equiv0(\bmod\; 23)$ and $P'(7)=14\not\equiv0(\bmod\; 23)$, we can just apply Hensel again. $\square$ $\textbf{Lemma 5:}$ $\forall k\in\mathbb{Z_+}\exists(x_k,y_k)\in\mathbb{F}_{2^k}^2$ such that $2^{k+1}\mid x_k^2+5y_k^2+2024$. $\textit{Proof.}$ Fix $y_k=2$ and consider the polynomial $P(x)=x^2+2044$. Since $2\nu_{2}(P'(2))<\nu_{2}(P(2))$, the strong version of Hensel's Lemma guarantees the existence of $x_k$. $\square$ $\textbf{Lemma 6:}$ $\nu_{11}(m)\le1$ $\textit{Proof.}$ Notice that $x^2+5y^2+2024\equiv0(\bmod\; 11)\implies x^2\equiv-5y^2(\bmod\; 11)\implies(xy^{-1})^2\equiv-5(\bmod\; 11)$ or $y\equiv0(\bmod\; 11)$, but $(\frac{-5}{11})=(\frac{-1}{11})(\frac{5}{11})=(-1)(\frac{11}{5})(-1)^{\frac{5-1}{2}\frac{11-1}{2}}=-(\frac{1}{5})=-1$, so the only option is $y\equiv0(\bmod\; 11)$ and then $x\equiv0(\bmod\; 11)$ too. So we get $x^2+5y^2\equiv0(\bmod\; 121)\implies x^2+5y^2+2024\equiv2024\not\equiv0(\bmod\; 121)$. $\square$ Putting together $\textbf{Lemmas 1, 2, 3, 4, 5, 6}$ and $\textbf{CRT}$, we have indeed verified that the solution set is that described above, so we are done. $\blacksquare$