Let us consider the principal value of the conditionally convergent infinite harmonic series
$$
\begin{align}
f(z)
&=\sum_{k=-\infty}^\infty\frac{1}{z+k}\\
&=\lim_{n\to\infty}\sum_{k=-n}^n\frac{1}{z+k}\tag{1a}\\
&=\lim_{n\to\infty}\frac1z+\sum_{k=1}^n\frac{1}{z-k}+\frac{1}{z+k}\tag{1b}\\
&=\frac1z+\sum_{k=1}^\infty\frac{2z}{z^2-k^2}\tag{1c}
\end{align}
$$
The series in $(1c)$ converges absolutely for all non-integer $z$.
Each of the terms in $(1c)$ is odd, so $f(-z)=-f(z)$.
The series in $(1a)$ shows that $f$ has a simple pole with residue $1$ at each integer.
$f$ has period $1$:
$$
\begin{align}
f(z+1)-f(z)
&=\lim_{n\to\infty}\sum_{k=-n}^n\frac{1}{z+k+1}-\frac{1}{z+k}\\
&=\lim_{n\to\infty}\frac{1}{z+n+1}-\frac{1}{z-n}\\
&=0\tag{2}
\end{align}
$$
$f(1/2)=0$:
$$
\begin{align}
f(1/2)
&=\lim_{n\to\infty}\sum_{k=-n}^n\frac{1}{k+1/2}\\
&=\lim_{n\to\infty}\frac{1}{n+1/2}\\
&=0\tag{3}
\end{align}
$$
Take the derivative of $f$:
$$
f'(z)=\lim_{n\to\infty}\sum_{k=-n}^n\frac{-1}{(z+k)^2}\tag{4}
$$
This series converges absolutely. and the terms monotonically go to $0$ as $|\Im(z)|\to\infty$.
Let's consider $if(iy)$ as $y\to\infty$. Using $(1c)$, we get
$$
\begin{align}
if(iy)
&=\frac1y+\sum_{k=1}^\infty\frac{2y}{y^2+k^2}\\
&=\frac1y+\sum_{k=1}^\infty\frac{2/y}{1+(k/y)^2}\tag{5}
\end{align}
$$
As $y\to\infty$, the summation in $(5)$ is a Riemann sum for the integral
$$
\int_0^\infty\frac{2\mathrm{d}x}{1+x^2}=\pi\tag{6}
$$
Thus, $if(iy)\to\pi$ as $y\to\infty$ and $if(iy)\to-\pi$ as $y\to-\infty$.
Since $f$ has period $1$ and $f'(z)\to0$ as $|\Im(z)|\to\infty$, it is evident that $f(z)\to-i\pi$ as $\Im(z)\to\infty$ and $f(z)\to i\pi$ as $\Im(z)\to-\infty$. This means that $f$ is bounded when away from the real axis.
The functions $f$ and $\pi\cot(\pi z)$ have the same poles, with identical
residues, and both are bounded when away from the real axis. Thus,
their difference is bounded for all $z$. Since their difference is
analytic and bounded, it must be constant. This difference is $0$ at
$1/2$, so it must be $0$ everywhere. Therefore, the principal value of
$$
\sum_{k=-\infty}^\infty\frac{1}{z+k}=\pi\cot(\pi z)\tag{7}
$$
for all $z$.
Combining $(1c)$ and $(7)$ yields
$$
\frac1z+\sum_{k=1}^\infty\frac{2z}{z^2-k^2}=\pi\cot(\pi z)\tag{8}
$$
Therefore,
$$
\begin{align}
\sum_{k=1}^\infty\frac{1}{k^2-z^2}
&=\frac{1}{2z^2}-\frac{\pi\cot(\pi z)}{2z}\\
&=\frac{1}{2z}\left[\frac1z-\pi\cot(\pi z)\right]\tag{9}
\end{align}
$$
