If ${27}^{x+1}=9^{x+3}=3^y$ , then $\left(x,y\right)$  is

${27}^{x+1}=9^{x+3}$

${\left(3^3\right)}^{x+1}={\left(3^2\right)}^{x+3}$

$3^{3x+3}=3^{2x+6}$

$3x-2x=6-3$

$x=3$

$9^{x+3}=3^y$

${\left(3^2\right)}^{3+3}=3^y$

${\left(3^2\right)}^6=3^y$

$3^{12}=3^y$

$y=12$

$\therefore \left(x,y\right)=\left(3,12\right)$