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} $