Intuitively this is simple and to prove the backwards direction: $\tau = 0 \Rightarrow \mathbf{b'}=0 \Rightarrow \mathbf{b} $ is constant.
Then letting $\gamma$ be the parametrisation of the curve and $s$ its arc-length, $\frac{d}{ds} (\gamma \cdot \mathbf{b}) = \dot{\gamma} \cdot \mathbf{b} + \gamma \cdot \mathbf{b'} = \dot{\gamma} \cdot \mathbf{b} = \mathbf{t} \cdot \mathbf{b} = 0 $
Hence $\gamma \cdot \mathbf{b}$ is constant and so the space curve is contained within the plane.
But how could I go about showing the opposite direction?