How to solve: $\frac{2^{n+1}}{n+1}=\frac{4+2^n}{3}$ $n$ is an integer variable satisfying $$\frac{2^{n+1}}{n+1}=\frac{4+2^n}{3}$$ How can I find $n$?

$\Large \frac{2^{n+1}}{n+1}=\frac{4+2^{n}}{3}$
this equation tells us that $n+1>0 \Rightarrow n>-1 $((i)we can't divide by zero (ii) RHS is +ve so LHS must also be +ve)
$\Large \frac{2^{n+1}}{n+1}=\frac{8+2^{n+1}}{6}=\frac{8}{6}+\frac{2^{n+1}}{6}$(multiply and divide RHS by $2$)
$\Large \frac{2^{n+1}}{n+1}-\frac{2^{n+1}}{6}=\frac{8}{6}$(difference of two terms is +ve)
this equation tells us that 
$\Large \frac{2^{n+1}}{n+1} > \frac{2^{n+1}}{6}\Rightarrow n+1<6 \Rightarrow n<5$
So the integer values to check for the answer are $0,1,2,3,4$
