Why are enthalpy of combustion and molar mass of organic compounds proportional?
The biggest contributor to the heat of combustion is the difference in covalent bond energy of reactants and products. The intermolecular forces that have to be overcome for combustion of liquids (with products all gases) are an order of magnitude lower (adapted from https://chemistry.stackexchange.com/a/125013):
$$
\begin{array}{lrr}
\hline
n\text{-Alkane} &amp; T_\mathrm{b}/\pu{K} &amp; \Delta_\mathrm{vap}H(T_\mathrm{b})/\pu{J mol-1}\\\hline
\text{Methane} &amp; 112 &amp; 8176\\
\text{Ethane} &amp; 184 &amp; 14640\\
\text{Propane} &amp; 231 &amp; 18832\\
n\text{-Butane} &amp; 273 &amp; 22390\\
n\text{-Pentane} &amp; 309 &amp; 26352\\
n\text{-Hexane} &amp; 342 &amp; 28850\\
n\text{-Heptane} &amp; 372 &amp; 31800\\
n\text{-Octane} &amp; 399 &amp; 33972\\
n\text{-Nonane} &amp; 424 &amp; 36910\\
n\text{-Decane} &amp; 447 &amp; 38750\\
\hline
\end{array}
$$

[OP] I know that the longer the chain is, the stronger the LDF get, so it is expected to see an increase in heat of combustion if molar mass increases.

Some of these alkanes are gases, so there are no dispersion forces to overcome. Also, you got the sign of the contribution wrong. If a reactant has higher LDF than the products (which have none), this would make the reaction less exothermic. So you would expect lower magnitudes for the larger alkanes, opposite of what is observed.
Instead, the heat of combustion scales with the number of carbons (see e.g. this table and graph) because the number of $\ce{C=O}$ and $\ce{O-H}$ bonds made (and the number of $\ce{C-C}$ and $\ce{C-H}$ bonds broken) also scales with the number of carbons.