  If the equation of the title has an integer solution with $k\ge2$, then
$m>10^{9.3\cdot10^6}$. This was the current best result and proved using a
method due to L. Moser (1953). This approach cannot be improved to reach the
benchmark $m>10^{10^7}$. Here we achieve $m>10^{10^9}$ by showing that
$2k/(2m-3)$ is a convergent of $\log2$ and making an extensive continued
fraction digits calculation of $(\log2)/N$, with $N$ an appropriate integer.
This method is very different from that of Moser. Indeed, our result seems to
give one of very few instances where a large scale computation of a numerical
constant has an application.
