$\newcommand{\R}{\mathbb R}$The roots are \begin{equation*} \begin{aligned} x_1&:=\frac{1}{2} \left(m-\frac1{\sqrt3}\sqrt{\frac{4 \sqrt[3]{2} m^4}{t}-5 m^2+2^{5/3} t}\right), \\ x_2&:=\frac{1}{2} \left(m+\frac1{\sqrt3}\sqrt{\frac{4 \sqrt[3]{2} m^4}{t}-5 m^2+2^{5/3} t}\right), \\ x_3&:=\frac{1}{2} \left(m-\sqrt{m^2-4 \left(\frac{\left(1+i \sqrt{3}\right) m^4}{3\ 2^{2/3} t}+\frac{2 m^2}{3}+\frac{\left(1-i \sqrt{3}\right) t}{6 \sqrt[3]{2}}\right)}\right), \\ x_4&:=\frac{1}{2} \left(m+\sqrt{m^2-4 \left(\frac{\left(1+i \sqrt{3}\right) m^4}{3\ 2^{2/3} t}+\frac{2 m^2}{3}+\frac{\left(1-i \sqrt{3}\right) t}{6 \sqrt[3]{2}}\right)}\right), \\ x_5&:=\frac{1}{2} \left(m-\sqrt{m^2-4 \left(\frac{\left(1-i \sqrt{3}\right) m^4}{3\ 2^{2/3} t}+\frac{2 m^2}{3}+\frac{\left(1+i \sqrt{3}\right) t}{6 \sqrt[3]{2}}\right)}\right), \\ x_6&:=\frac{1}{2} \left(m+\sqrt{m^2-4 \left(\frac{\left(1-i \sqrt{3}\right) m^4}{3\ 2^{2/3} t}+\frac{2 m^2}{3}+\frac{\left(1+i \sqrt{3}\right) t}{6 \sqrt[3]{2}}\right)}\right), \end{aligned} \end{equation*} where \begin{equation*} t:=\sqrt[3]{3 \sqrt{3} \sqrt{L \left(27 L-4 m^6\right)}-27 L+2 m^6}. \end{equation*}
Clearly, $x_1\in\R\iff x_2\in\R$, $x_3\in\R\iff x_4\in\R$, and $x_5\in\R\iff x_6\in\R$, since $m$ is real.
Using the Mathematica Reduce command, we get that
$x_3$ and $x_5$ are never real and hence $x_4$ and $x_6$ are never real (if $m$ and $L$ are positive integers)
$x_1\in\R\iff x_2\in\R\iff m\ge z_2$, where $z_2$ is the positive root $z$ of the equation $9z^6=64L$.
Thus, only $x_1$ and $x_2$ can ever be real roots of your sextic equation, and they are actually real (for given positive integers $m$ and $L$) iff \begin{equation*} L\le\frac9{64}\,m^6. \end{equation*}