$x$ , $y$ , and $x$ -ponents

From equation,

$\begin{aligned}\dfrac{4^x}{2^{x+y}}=&\ 8\\\dfrac{2^{2x}}{2^{x+y}}=&\ 2^3\\2^{2x}=&\ 2^3(2^{x+y})\\\color{#624F41}2^{2x}=&\ \color{#624F41}2^{3+x+y}\\2x=&\ 3+x+y\\\color{#3D99F6}{x=}&\ \color{#3D99F6}{3+y}\end{aligned}$

Instead $\color{#3D99F6}x$ with $\color{#3D99F6}{3+y}$ in equation $\dfrac{9^{\color{#3D99F6}x+y}}{3^{5y}}=243$ that is,

$\begin{aligned}\dfrac{9^{\color{#3D99F6}{3+y}+y}}{3^{5y}}=&\ 243\\\dfrac{9^{3+2y}}{3^{5y}}=&\ 3^5\\\dfrac{3^{6+4y}}{3^{5y}}=&\ 3^5\\(3^{6+4y})(3^{-5y})=&\ 3^5\\\color{#624F41}3^{6-y}=&\ \color{#624F41}3^5\\6-\color{#D61F06}y=&\ 5\Rightarrow \color{#D61F06}{y=1}\end{aligned}$

instead $\color{#D61F06}y$ with $\color{#D61F06}1$ in equation $\color{#3D99F6}{x=3+y}$ that is, $\color{teal}x=3+\color{#D61F06}1=\color{teal}4$

therefore, $\color{teal}x+\color{#D61F06}y=\color{teal}4+\color{#D61F06}1=\boxed{5}$