You can write
$ \frac{1}{n^2 - m^2} = \frac{1}{2n} \left\lbrace \frac{1}{m+n} - \frac{1}{m-n} \right\rbrace . \quad (1)$
Now if we sum up both sides over all odd $m \ne n ,$ taking into account that $n$ is odd, lots of cancelling goes on and we obtain
$\sum_{m \ne n} \frac{1}{n^2 - m^2} = -\frac{1}{4n^2}.$
At first sight it appears the cotangent identity could be useful but it's not actually needed.
As a numerical check try summing the following with wolframAlpha
$1/24 + 1/16 + \sum_{k=3}^\infty 1/(5^2 - (2k+1)^2),$
you will see that it is $-1/100,$ as expected.
Or try this:
$1/48 + 1/40 + 1/24 + \sum_{k=4}^\infty 1/(7^2 - (2k+1)^2).$
You will get $-1/196.$
EDIT: To clarify the cancellation taking place when we sum the RHS of $(1).$
We have
$\sum_{m \ne n, \,\, m \textrm{ odd} } \left\lbrace \frac{1}{m+n} - \frac{1}{m-n} \right\rbrace = \sum_{m \ne n, \,\, m \textrm{ odd} } \left\lbrace \frac{1}{n+m} + \frac{1}{n-m} \right\rbrace $
$= \left\lbrace \left( \frac{1}{n+1} + \frac{1}{n-1} \right) + \left( \frac{1}{n+3} + \frac{1}{n-3} \right) + \left( \frac{1}{n+5} + \frac{1}{n-5} \right) + \cdots + \left( \frac{1}{2n-2} + \frac{1}{2} \right) \right\rbrace $
$+ \left( \frac{1}{2n+2} - \frac{1}{2} \right) + \left( \frac{1}{2n+4} - \frac{1}{4} \right) + \left( \frac{1}{2n+6} - \frac{1}{6} \right) + \cdots $
and rearranging all the terms in the braces
$= \left\lbrace \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2n-2} \right\rbrace + \left( \frac{1}{2n+2} - \frac{1}{2} \right) + \left( \frac{1}{2n+4} - \frac{1}{4} \right) + \cdots $
$=\left\lbrace \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2n-2} \right\rbrace - \left\lbrace \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2n} \right\rbrace = - \frac{1}{2n}$
and hence the result.