This becomes easier by converting to an integral to get that weird linear term out of the denominator. Jack gave the quick version in the comments, but here's the more detailed explanation.
First, introduce an integral to convert from a zeta-like series to a power series:
\begin{multline}
\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}\sum_{n=0}^\infty \frac{300}{2^nk+5} = 300\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}\sum_{n=0}^\infty\int_0^1 x^{2^nk+4}dx
\\= 300\int_0^1x^4\left[\sum_{n=0}^\infty\sum_{k=1}^\infty (-1)^{k-1}\frac{x^{2^nk}}{k}\right] dx = 300\int_0^1x^4\left[\sum_{n=0}^\infty\ln\left(1+x^{2^n}\right)\right] dx,
\end{multline}
where we used the series expansion $\ln(1+z) = \sum_{k=0}^\infty (-1)^{k-1}z^k/k$ to do the $k$ summation.
Next, evaluate the $n$ series using
$$
\sum_{n=0}^\infty \ln(1+x^{2^n}) = \ln\left[\prod_{n=0}^\infty \left(1+x^{2^n}\right)\right] = \ln\left[\sum_{m=0}^\infty x^m\right] = \ln\left(\frac{1}{1-x}\right) = -\ln(1-x).
$$
The identity with the infinite product can be seen by considering expanding each exponent $m$ as sums of powers of $2$. So now we've got our sum in terms of a nice integral,
$$
\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}\sum_{n=0}^\infty \frac{300}{2^nk+5} = -300\int_0^1x^4\ln(1-x)dx
$$
Last step is to use the identity $\int_0^1 x^n\ln(x)dx =-(n+1)^{-2}$ to get
$$
-300\int_0^1x^4\ln(1-x)dx = -300\int_0^1(1-x)^4\ln(x)dx = 300\left(1 - \frac{4}{4} + \frac{6}{9}-\frac{4}{16}+\frac{1}{25}\right) = 137
$$
So, in conclusion,
$$
\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}\sum_{n=0}^\infty \frac{300}{2^nk+5} =137
$$
