As was pointed out in the comments above, one has to use relativistic mechanics to talk meaningfully about massless particles; you can't just write $F = ma$ and expect it to work. And, indeed, it turns out that we can come up with a reasonable question of motion for a charged massless particle if we use the machinery of relativistic dynamics.
The equation of motion of a massive charged relativistic particle is
$$
\frac{d p^\mu}{d\tau} = \frac{q}{m} F^{\mu}{}_\nu p^\nu, \tag{1}
$$
where $m$ is the particle's rest mass, $q$ is its charge, $p^\mu$ is its four-momentum, $F^{\mu \nu}$ is the field strength tensor, and $\tau$ is the proper time along the (massive) particle's world-line. In particular, $\tau$ serves mainly as a parameter that traces out the particle's world-line through spacetime.
If we multiply both sides by $m$ and introduce a new parameter $\lambda = \tau/m$, it turns out that the above equation is equivalent to
$$
\frac{d p^\mu}{d \lambda} = q F^\mu {}_\nu p^\nu. \tag{2}
$$
What's more, under this parameterization we have
$$
\frac{d x^\mu}{d \lambda} = m \frac{d x^\mu}{d \tau} = p^\mu. \tag{3}
$$
The new parametrization (2) has a perfectly well-behaved limit as $m \to 0$; in fact, this is a conventional parametrization for the worldlines of massless particles. Defining our $x$-direction to be the direction of the field, we have $F^{01} = - F^{10} = E_x$, and so the equations of motion are
\begin{align*}
\dot{p}^t &= q E_x p^x & \dot{p}^x &= q E_x p^t & \dot{p}^y = \dot{p}^z = 0 \tag{4}
\end{align*}
where dots denote differentiation with respect to $\lambda$. This can then be solved (in principle) for the four-momentum $p^\mu$ as a function of $\lambda$. If desired, one can then find the trajectories $t(\lambda)$, $x(\lambda)$, etc. in parametric form, and (in principle) invert the first of these to obtain $x(t)$, $y(t)$ and $z(t)$.
This means that the energy ($p^t$) and $x$-momentum ($p^x$) of a massless charged particle will increase as it travels. But you can also show from the equations of motion (4) (try it!) that the quantity $p_\mu p^\mu = -(p^t)^2 + (p^x)^2 + (p^y)^2 + (p^z)^2$ is constant with respect to $\lambda$. So these equations ensure that a massless charged particle (with $p_\mu p^\mu = - m^2 = 0$) stays massless as it travels; and that means that it always travels at the speed of light, even as its energy and momentum change.
