Squaring and rearranging: $$\sqrt{x}+\sqrt{y}=\sqrt{x+2023}$$$$\implies x+2\sqrt{xy}+y=2023+x$$$$\implies 4xy=(2023-y)^2$$If $y$ is odd, then $4\mid(2023-y)^2$. So for $x$ to be an integer, we only need $y\mid(2023-y)^2$ or equivalently $y\mid2023^2$. We note that $y$ must satisfy $0\le y\le 2023$, due to the square root in the original equation. Since $2023=7\cdot17^2$, we compute the factors of $2023^2$ not greater than $2023$, giving $y\in\{1,7,49,119,289,833,2023\}$. So the only pairs $(x,y)$ that are solutions to our equation are: $(0,2023)$, $(425,833)$, $(2601,289)$, $(7616,119)$, $(19881,49)$, $(59177,17)$, $(145152,7)$, and $(1022121,1)$.

Rainmaker2627 wrote: Squaring and rearranging: $$\sqrt{x}+\sqrt{y}=\sqrt{x+2023}$$$$\implies x+2\sqrt{xy}+y=2023+x$$$$\implies 4xy=(2023-y)^2$$If $y$ is odd, then $4\mid(2023-y)^2$. So for $x$ to be an integer, we only need $y\mid(2023-y)^2$ or equivalently $y\mid2023^2$. We note that $y$ must satisfy $0\le y\le 2023$, due to the square root in the original equation. Since $2023=7\cdot17\cdot289$, we compute the factors of $2023^2$ not greater than $2023$, giving $y\in\{1,7,49,119,289,833,2023\}$. So the only pairs $(x,y)$ that are solutions to our equation are: $(0,2023)$, $(425,833)$, $(2601,289)$, $(7616,119)$, $(19881,49)$, $(59177,17)$, $(145152,7)$, and $(1022121,1)$. $2023=7*17^2$

$(2023+2x-y)^2=(2\sqrt{x^2+2023x})^2$ Expanding this gives us $(y-2023)^2=(2\sqrt{xy})^2$ LHS is integer, so $xy$ is a perfect square. $$x=\frac{k^2}{y}$$Putting this into $(y-2023)^2=(2\sqrt{xy})^2$ gives $y=2k+2023\Rightarrow x=\frac{k^2}{y}=\frac{k^2}{2k+2023}$ After some work,you should get $4x=2k-2023+\frac{2023^2}{2k+2023}$ $$2k+2023\in\{ 1;7;17;7\cdot17;17^2;7\cdot17^2 \}\Rightarrow y\in\{ 1;7;17;7\cdot17;17^2;7\cdot17^2 \}$$$$(x;y)=(1011^2;1)(\frac{1008^2}{7};7)(\frac{1003^2}{17};17)(\frac{952^2}{7\cdot17};119)(\frac{864^2}{17^2};289)(0;2023)$$