Omitting the case $m = n = 0$, if,
$$ \sum_{m,n = - \infty}^{\infty} \frac{(-1)^m}{m^2 + 58 n^2} = - \frac{2\pi \ln\big(\tfrac{5 + \sqrt {29}}{\sqrt2}\big)}{\sqrt {58}} $$
as in this post, then is,
$$ \sum_{m,n = - \infty}^{\infty} \frac{(-1)^m}{m^2 + mn+ 41 n^2} = - \frac{2\pi \ln\big(\beta\big)}{\sqrt {163}} $$
for some algebraic number $\beta$? If yes, then what is $\beta$?
P.S. Incidentally, we have the nice approximation,
$$e^{\pi\sqrt{58}} \approx \Big(\tfrac{5 + \sqrt {29}}{\sqrt2}\Big)^{12} +23.999999988776\dots$$
and the "excess" near $24$ has to do with the Dedekind eta function.
The key here is a result which goes by the name of Kronecker's second limit formula. Using the formulation given in Wikipedia it can be proved that the desired sum in question is equal to $$-\frac{2\pi\log|2g^{4}(q)|} {\sqrt{163}} $$ where $q=\exp(\pi i\tau), 2\tau=1+i\sqrt{163}$ and $$g(q) =2^{-1/4}q^{-1/24}\prod_{n=1}^{\infty}(1-q^{2n-1})$$ With some manipulation it can be shown that the above sum is equal to $$-\frac{\pi\log(2G_{163}^{4})}{\sqrt{163}}$$ The value of $$G_{163}=\frac{6+\sqrt[3]{135-3\sqrt{489}}+\sqrt[3]{135+3\sqrt{489}}} {3\sqrt[4]{2}} $$ is (not so) well known and the calculations explained above can be performed with reasonable amount of labor to obtain a closed form for the sum in question.