I'm assuming that "$[\cdots]$" is the floor function, sometimes written "$\lfloor \cdots \rfloor$".
Since $m$ and $n$ are integers we have
$\Gamma(n+1) = n! \qquad \Gamma(m+1) = m! \qquad \Gamma(m+1-n) = (m-n)!$
This allows us to rewrite the left-hand side of the inequality as
$ 2^{m-k} \frac{\Gamma\left(\left[\frac{m+1-k}{2}\right]\right)}{\Gamma\left(\left[\frac{m+2-k}{2}\right]\right)} \cdot \frac{n!(m-n)!}{m!} = 2^{m-k} \frac{\Gamma\left(\left[\frac{m+1-k}{2}\right]\right)}{\Gamma\left(\left[\frac{m+2-k}{2}\right]\right)} \cdot \frac{1}{\binom{m}{n}}. $
Now if $m-k$ is even then $m+2-k$ is even, which implies that $(m+2-k)/2$ is an integer and
$ \left[\frac{m+2-k}{2}\right] = \frac{m+2-k}{2} = \frac{m-k}{2} + 1. $
This tells us that
$ \Gamma\left(\left[\frac{m+2-k}{2}\right]\right) = \Gamma\left(\frac{m-k}{2} + 1\right) = \frac{m-k}{2} \,\Gamma\left(\frac{m-k}{2}\right). $
Then since
$ \frac{m+1-k}{2} = \frac{m+2-k}{2} - \frac{1}{2}, $
we have
$ \left[\frac{m+1-k}{2}\right] = \frac{m+2-k}{2} - 1 = \frac{m-k}{2}, $
which gives us
$ \Gamma\left(\left[\frac{m+1-k}{2}\right]\right) = \Gamma\left(\frac{m-k}{2}\right). $
Some cancellation happens in the inequality and we are left with
$ \frac{2^{m+1-k}}{(m-k)\binom{m}{n}} \leq \sqrt{\pi}. $
After multiplying by the denominator and taking logarithms this is equivalent to
$ k \geq m+1 - \log_2\left[\sqrt{\pi}(m-k)\binom{m}{n}\right] \qquad \text{if } m-k \text{ is even.} $
If $m-k$ is odd then similar considerations reveal that
$ \Gamma\left(\left[\frac{m+1-k}{2}\right]\right) = \Gamma\left(\left[\frac{m+2-k}{2}\right]\right), $
which leaves the inequality in the form
$ \frac{2^{m-k}}{\binom{m}{n}} \leq \sqrt{\pi}. $
By again multiplying by the denominator and taking logarithms we see that this is equivalent to
$ k \geq m - \log_2\left[\sqrt{\pi}\binom{m}{n}\right] \qquad \text{if } m-k \text{ is odd.} $