A different perspective on an algebraic solution: you know that for negative $x$, $2^x\lt 1$ but $1-x \gt 1$ (since the latter is $1+(-x)$ and $-x \gt 0$); contrariwise, for positive $x$, $2^x\gt 1$ but $1-x \lt 1$. This means that the only possibility for a solution is $x=0$, and of course by quick algebra that does in fact work.