Well, using WolframAlpha, one has the identity

\begin{equation} \sum_{k=0}^{\infty} \frac{1}{k^2 + a^2} = \frac{\pi a \coth (\pi a) + 1}{2a^2} \end{equation}

which, after separating the sum into odd/even contributions and factorizing the terms, leads to :

\begin{equation} \begin{array}{rcl} \sum_{k=0}^{\infty} \frac{1}{(2k+1)^2 + m^2} &=& \frac{\pi}{2} (\coth (\pi m) - \frac{\coth (\frac{\pi m}{2})}{2}) - \frac{1}{m^2} \\ &=& \frac{\pi}{4 c}- \frac{1}{m^2} \end{array} \end{equation}

where $c := \coth( \frac{\pi m}{2}) $, thanks to the addition formula for $\coth$.

Thus, your sum is equal to \begin{equation} \frac{\pi}{4} \sum_{m=1}^{\infty} \frac{(-1)^m}{\coth(\frac{\pi m}{2})} - \sum_{m=1}^{\infty} \frac{(-1)^m}{m^2} \end{equation}

Okay, that's not a complete proof of the equality, but at least $\pi$ appeared.

The second sum is computed by splitting again in even/odd contributions and gives you the $\frac{\pi^2}{16}$ contribution (that's kind of a twist of the usual $\zeta(2)$ value).

Well, using WolframAlpha, one has the identity

\begin{equation} \sum_{k=0}^{\infty} \frac{1}{k^2 + a^2} = \frac{\pi a \coth (\pi a) + 1}{2a^2} \end{equation}

which, after separating the sum into odd/even contributions and factorizing the terms, leads to :

Edit : corrected after Hjalmar's remark

\begin{equation} \sum_{k=0}^\infty\frac 1{(2k+1)^2+m^2}=\frac{\pi\tanh(\pi m/2)}{4m} \end{equation}