Mass-luminosity relation for a fully convective star

The bit you are missing is that the radius of the star is determined by the opacity.

Hydrostatic equilibrium yields $$P_R = frac{GM}{R^2}int rho dr,$$ where $P_R$ is the pressure at the radius where light can escape. This in turn is related to the opacity $kappa$ by $$int kappa rho dr = bar{kappa}int rho dr = 1$$ Thus $$P_R = frac{GM}{R^2bar{kappa}} propto frac{GM}{R^2 rho^a T_{rm eff}^b}, tag*{(1)}$$ where I have substituted in $bar{kappa}propto rho_R^a T_{rm eff}^b$.

With this equation you also have $$ L = 4pi R^2 sigma T_{rm eff}^4 tag*{(2)}$$ $$ L propto M rho T^c tag*{(3)}$$

From the virial theorem and perfect gas law you also know that the interior $T propto M/R$ and interior $rho propto M/R^3$.

You thus have 6 variables $L, M, R, T_{rm eff}$, $P_R$, $rho_R$ and 3 equations. To eliminate all but $L$ and $M$ you need another.

This comes from the interior polytropic equation of state $P propto rho^{gamma}$ for a fully convective star (a polytrope with $gamma = 5/3$ for a fully convective star). However, this proportionality only applies to a particular star - the constant of proportionality in fact depends on the mass and radius of the star in question. Here is not the place to repeat textbooks that deal with polytropes, but it can be shown that (e.g. here) $$Prho^{-gamma} propto P^{1-gamma} T^{gamma} propto M^{2-gamma} R^{3gamma -4}$$ and thus near the surface $$P_R^{1 -gamma} T_{rm eff}^{gamma} propto M^{2 -gamma} R^{3gamma-4} tag*{(4)}$$

You then solve to get $L$ only as a function of $M$ and substitute in $gamma=5/3$ and your favourite values for $a$, $b$ and $c$.