We will assume that $x,m > 0$. Our results show (unless there is some mistake in the case analysis) that the only solutions (up to a value of $x$) are given by $ \frac{1}{10C} = \frac{1}{14C} + \frac{1}{35C} $ and the exceptional solution $ \frac{1}{3} = \frac{1}{\lfloor 2\cdot 2.4\rfloor} + \frac{1}{\lfloor 5\cdot 2.4\rfloor} = \frac{1}{4} + \frac{1}{12}. $
As user9176 mentions, intuitively it is obvious that we can assume that $x$ is of the form $A + B/10$, where $B \in \{0,\cdots,9\}$. We then have $\lfloor 2x \rfloor = 2A + \alpha, \; \lfloor 5x \rfloor = 5A + \beta, $ where $\alpha,\beta$ depend on $B$ and take on the six values $(0,0);(0,1);(0,2);(1,2);(1,3);(1,4). $ We can thus rewrite the equation as $ (2A + \alpha) (5A + \beta) = m(7A + \alpha + \beta), $ from which we can form a quadratic $ 10A^2 - (7m - 5\alpha - 2\beta)A - ((\alpha+\beta)m - \alpha\beta). $ The discriminant of the quadratic is $ (7m - 5\alpha - 2\beta)^2 + 40 ((\alpha+\beta)m - \alpha\beta). $ For the equation to have integer solutions, we need the discriminant to be a perfect square. Let's see what that implies for each of the six pairs $(\alpha,\beta)$.
The first case, $(0,0)$, doesn't actually require the theory. The equation simply reads $10A^2 = 7mA$, or $10A = 7m$, which implies that $7|A$. On the other hand, if $A = 7C$, then we can recover $m = 10C$. We get the parametric solution $ m = 10C, x \in [7C,7C+0.2). $
Next, consider $(0,1)$. The discriminant is $ (7m-2)^2 + 40m. $ The squares following $(7m-2)^2$ are $(7m-2)^2 + \{14m - 3, 28m - 4, 42m - 3, 56m\}.$ Comparing coefficients, we see that the discriminant cannot be a perfect square.
Next, consider $(0,2)$. The discriminant is $ (7m-4)^2 + 80m. $ The squares following $(7m-4)^2$ are $(7m-4)^2 + \{14m-7, 28m-12, 42m-15, 56m-16, 70m-15, 84m-12, 98m-7\}.$ Comparing coefficients, the only solution is $m = 3$. The quadratic now reads $ 10A^2 - 17A - 6 = 0 = (10A + 3)(A - 2). $ The only integral solution is $A = 2$, and we get the solution $ m = 3, x \in [2.4,2.5).$
Next, consider $(1,2)$. The discriminant is $ (7m-9)^2 + 120m - 80. $ The squares following $(7m-9)^2$ are $\begin{align*} (7m-9)^2 + \{&14m-17,28m-32,42m-45,56m-56,70m-65,\\&84m-72,98m-77,112m-80,126m-81,140m-80,154m-77\}. \end{align*}$ This time there are no integral solutions.
Next, consider $(1,3)$. The discriminant is $ (7m-11)^2 + 160m - 120. $ The squares following $(7m-11)^2$ are $\begin{align*} (7m-11)^2 + \{&14m-21, 28m-40, 42m-57, 56m-72, 70m-85, \\ &84m-96, 98m-105, 112m-112, 126m-117, 140m-120, \\ &154m-121, 168m-120, 182m-117\}. \end{align*}$ Again there are no integral solutions.
Finally, consider $(1,4)$. The discriminant is $ (7m-13)^2 + 200m - 160. $ The squares following $(7m-13)^2$ are $\begin{align*} (7m-13)^2 + \{& 14m-25, 28m-48, 42m-69, 56m-88, 70m-105, \\& 84m-120, 98m-133, 112m-144, 126m-153, 140m-160, \\& 154m-165, 168m-168, 182m-169, 196m-168, 210m-165, \\& 224m-160, 238m-153\}. \end{align*}$ Once again there are no integral solutions.