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}