Uniqueness conjecture for a Stirling-number Diophantine equation

For integers $n$ and $m$, consider the Diophantine equation

$$\binom{n}{4}+10\binom{n}{5}+15\binom{n}{6}=\binom{m}{2}.$$

The displayed Stirling-number identity shows that this equation arises from equating $\genfrac\{\}{0pt}{}{n}{n-3}$ and $\genfrac\{\}{0pt}{}{m}{m-1}$. Uniqueness conjecture. The unique solution with $n\geq 6$ is

$$(n,m)=(14,364).$$

The authors found this as the unique solution in the computational range $6\leq n\leq 4.6\cdot 10^7$, while uniqueness for all $n\geq6$ remains open.