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}