Assuming $c=1$, $v=\frac{\omega}{k}=\frac{\sqrt{k^2+m^2}}{k}>1$, for $m \neq 0$. Why is it not an issue that this $v$ is greater than the speed of light?
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No. The physical speed is the group velocity $$ v_g= \frac{\partial \omega}{\partial k} = \frac{k}{\sqrt{k^2+m^2}} <1. $$ The phase velocity $$ v_\phi = \frac{\omega}{k} $$ is not relevant.
mike stone
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