To directly answer your question, a particle having zero rest mass implies that it must be moving with speed $c$, and objects which move at $c$ in one frame must do so in all frames. On the other hand, if a particle is moving at speed $v=c$ then its energy must be equal to $pc$, which implies its mass is equal to zero. In that sense, $m=0$ implies, and is implied by, $v=c$. In the following, I'll try to explain this in more depth, and then mention the (largely unfashionable) concept of relativistic mass.
Here is my favorite interpretation of rest mass in special relativity. The general energy-momentum relationship for a particle is
$$E=\sqrt{p^2c^2 + m^2c^4}$$
If $m=0$, then $E = pc$, and the relationship between $E$ and $p$ becomes linear. If $m\neq 0$, then the relationship is nonlinear, but becomes approximately linear for values of $p$ which are large compared to $mc$.
This is a plot of energy vs momentum for various values of $m$ (in natural units, $c=1$), with the non-relativistic $\frac{p^2}{2m}$ approximation superposed on top with dotted lines. As you can see, the mass of the particle defines a particular momentum scale $p = mc$, below which the nonrelativistic approximation is good and above which the energy/momentum relationship is essentially linear.
The smaller the mass of the particle, the smaller the range of momenta for which the particle could be considered nonrelativistic; if the particle has zero mass, then it is relativistic for all values of its momentum, as is the case for the photon.
At the same time, one can express the velocity with which a particle moves as
$$\mathbf v = \frac{\mathbf p c^2}{E}= \frac{\mathbf pc^2}{\sqrt{p^2 c^2+m^2c^4}} = c \frac{\mathbf p}{\sqrt{p^2+m^2c^2}}$$
Here is a plot of the magnitude of $\mathbf v$ against the magnitude of $\mathbf p$, again for various masses.
Just as before, the mass defines a cutoff. This time, for $p<mc$ we find that $v\approx p/m$, while for $p>mc$ we find that $v \approx c$. The smaller $m$ is, the smaller the range of momenta for which $p=mv$ is a good approximation. If the particle has zero mass, then $p=mv$ is invalid for all $p$, and we have simply that $v=c$.
The relativistic mass, on the other hand, arises as a desperate attempt to hold on to the expression $\mathbf p = m\mathbf v$. Inverting the expression in the last section,
$$\mathbf p = \frac{E}{c^2}\mathbf v $$
The quantity $\frac{E}{c^2}$ is defined to be the relativistic mass $m_r$. Note that $\frac{E}{c^2} = \sqrt{\frac{p^2}{c^2} + m^2}$, so if $m\neq 0$ this can be written
$$\frac{E}{c^2} = m\sqrt{1+\left(\frac{pc}{m}\right)^2} \equiv \gamma m$$
and so in that case,
$$ \mathbf p = m_r \mathbf v = \gamma m \mathbf v$$
Over the past 100 years, the physics community has largely decided that the concept of relativistic mass is more trouble than it's worth. It doesn't actually yield any useful insights - if anything, it obscures the fact that reality is fundamentally relativistic - so it has fallen out of favor as a concept.
