Let us say that the smallest divisor of $a$ is $k$ greater than $1$. This means that $a+b=(k+1)b$. In order for this to be a power of $2$, both $k+1$, and $b$ are powers of $2$. If $k=2$, then $k+1=3$ which does not work. If $k$ is and odd prime, then $k+1$ could be a power of $2$. However, this means that $2$ is not a divisor of $a$ meaning that $b$ is odd meaning that this does not work.

$\color{blue} \boxed{\textbf{SOLUTION}}$ First, Let, $a=2n \implies b=n \implies (a+b)=3n$ is not equal $2^k$ So, Let, $a=pk$ where $p$ is the smallest prime divisor of $a,$ $a$ is odd, we have $k \in$ odd as $p \in$ odd. $\implies b=k, (a+b)=pk+k=k(p+1)$ But $k$ is not equal to $2^m, m \in \mathbb Z$ So, $(a+b)$ is not equal to $2^m, m \in \mathbb Z \blacksquare$