Since $\frac{15x-7}{5}$ is required to be an integer, it is of the form $\frac{2+5j}{15}$ for $j$ an arbitrary integer. Thus, the equation to solve becomes $ \left\lfloor\frac{6\big(\frac{2+5j}{15}\big)+5}{8}\right\rfloor = \frac{15\big(\frac{2+5j}{15}\big)-7}{5}\qquad\text{which simplifies to}\quad \left\lfloor\frac{\frac{4}{5}+2j+5}{8}\right\rfloor = j-1 $ Now observe that the function on the right hand side increases quicker than the function on the left hand side. So we can solve the question by finding a maximal solution to the inequality $ j-1\leq\left\lfloor\frac{\frac{4}{5}+2j+5}{8}\right\rfloor $
A warning: But here we should be careful, because a maximal solution to this inequality is not necessarily a solution to our equality. Also, there might be more solutions than just the maximal solution $j_0$ to this inequality. But since the left hand side and the right hand side are linear in $j$, the set of solutions forms an interval in $\mathbb{Z}$. This means that to give all the solutions, it suffices to find the largest integer below $j_0$ for which the inequality is strict.
Note that for any $n\in\mathbb{Z}$ and $a\in\mathbb{R}$, the inequality $n\leq\lfloor a\rfloor$ holds if and only if $n\leq a$ holds. This helps, since now we can continue to compute $ \begin{split} j-1 \leq \left\lfloor\frac{\frac{4}{5}+2j+5}{8}\right\rfloor & \Leftrightarrow j-1 \leq \frac{\frac{4}{5}+2j+5}{8}\\ & \Leftrightarrow 8j-8\leq \frac{4}{5}+2j+5\\ & \Leftrightarrow 10j\leq 23\\ & \Leftrightarrow j\leq 2. \end{split} $ Therefore, we get the maximal solution for the equation by substituting $j=2$ in $x=\frac{2+5j}{15}$, i.e. $x=\frac{4}{5}$. This is indeed a solution to the equality. As we have noted before, $\frac{4}{5}$ is not necessarily the only solution. Indeed, it is easy to check that $j=1$ also gives a solution to the equation. But for $j\leq 0$ the inequality is strict, so the only solutions to the equation are $x=\frac{4}{5}$ and $x=\frac{7}{15}$.
Here is a link to the answer to another question, where the OP had to deal with the floor function. Maybe you'll find it useful.