jmc

Let $2^x=a$ and $3^y=b$. Since $2^{x+2}=2^2(2^x)$ and $3^{y+1}=3(3^y)$, the equations become

$$\begin{array}{rl}a+b& =5,\\ 4a+3b& =18.\end{array}$$Multiplying the first equation by $3$ and subtracting it from the second equation, we find $a=\boxed{3}$ and $b=2$. Plugging these into the original equations, we find this works.