Applying a time delay with Z-Transform
There's a lot of confusion in your question. You refer to "a channel $H(s)$", then to "my signal $H(s)$", so for the sake of clarity, let's properly define:


Input signal - continuous time $f(t)$, impulse sampled $f[n]=f(n\Delta T)$, Fourier transform of continuous time signal $F(s)$, Z-transform of impulse sampled signal $F(z)=F(e^{s\Delta T})$.
Continuous time channel - impulse response $h(t)$, transfer function (Laplace transform of impulse response) $H(s)$.
Discrete time channel - discrete impulse response $h[n]$, transfer function (Z-transform of discrete impulse response) $F(z)$.


Note the definitions above follow a convention:


time-domain signals/functions are lower case ($f, h$); frequency-domain signals/functions are upper case ($F, H$).
arguments of continuous signals/functions are enclosed in parentheses $(t)$; arguments of discrete signals/functions are enclosed in brackets $[n]$.


Then, you write:


  (...) z-transform for my channel but am unsure how I can go about the convolution with my sampled signal which is just an array of numbers


It seems you do not understand the basics of LTI systems. Note that $H(s)$ and $H(z)$ are frequency-domain and $f(t)$ and $f[n]$ are time-domain.

Convolution makes sense when the two convolved functions are in the same exact domain (continuous/discrete time/frequency), although for the usual purposes the two convolved functions should both be in the continuous/discrete time domain.

If you're dealing with a discrete-time signal ("signal which is just an array of numbers"), then you might want to convolve $f[n]$ with $h[n]$. If you already have $H(z)$, then an inverse Z-transform on $H(z)$ would produce $h[n]$.