$x-1$ and $x+1$ are $2$ apart

Powers of $2$ (two columns so that numbers above each other are $2$ powers apart):

$2$ $4$

$8$ $16$ $4$ and $16$ are $12$ apart

$32$ $64$ numbers are getting too far apart

$16-4=12$

$2^4-2^2=12$

$x=3$

$2^{x+1}=2^x\times2^1=2^x\times2$ and $2^{x-1}=2^x\times2^{-1}=2^x\times\frac12$.

Substitute then factorise: $$\begin{align} 2^{x+1}-2^{x-1} =&12\\

\Rightarrow2^x\times2-2^x\times\tfrac12=&12\\

\Rightarrow 2^x\left(2-\tfrac12\right)=&12\\

\Rightarrow 2^x\times\tfrac32=&12\\

\Rightarrow 2^x\times3=&12\times2\\

\Rightarrow 2^x=&24\div3\\

\Rightarrow 2^x=&8\end{align}$$ So $x=3$, since $2^3=8$.

You can find more short problems, arranged by curriculum topic, in our short problems collection.