So this is the proof I have:
If $p(z)$ is a non-constant polynomial, then there exists a $z \in \Bbb Z$ such that $p(z) = 0$.
Let $p(z) = z^n + a_{n-1}z^{n-1} +a_{n-2} z^{n-2} + ... + a_0$ where $n \in \Bbb N$ and $a_0, a_1, ..., a_{n-1} \in \Bbb C$. $\lim\limits_{z \to \infty} p(z) = z^n[1 + \frac{a_{n-1}}{z} + \frac{a_{n-2}}{z^2} + ... + \frac{a_0}{z^n}] = \infty$, so if $p(z)\neq 0$ for all $z \in \Bbb C$, then define $f(z) = \frac{1}{p(z)}$ so $\lim\limits_{z \to \infty} f(z) = 0$.
Then there exists an $N > 0$ such that $\lvert f(z) \rvert \leq 1$ for all $\lvert z \rvert > N$.
I don't understand how this bolded line is achieved or how it follows from the limit.
Continuing the proof:
Take the closed disk $\lvert z \rvert \leq N$. Since this disk is compact and $f$ is continuous, there exists an $M > 0$ such that $\lvert f(z) \rvert \leq M$ for all $\lvert z \rvert \leq N$.
I don't understand what the disk has to do with anything, or how these inequalities follow from that disk, or why we are re-using the letter $N$ again.
Finishing the proof:
Since $f(x)$ is bounded and entire, by Liouville $f(x)$ is a constant so $p(z)$ is a constant which is a contradiction.
I understand this ending but I don't understand the two previous lines.
Here I've collected the comments above.
So as stated we want to use Liouville's theorem on the function $f$. Thus we need to make sure that $f$ is bounded on all of $\mathbb C$. We do this in two steps.
1) Note that since $f$ has no zeros, and that $\lim_{z \to \infty}f(z)=0$. From the definition of limits we have that: $$\forall \epsilon >0 \, \exists N(\epsilon) >0\quad \mbox{such that } |z|>N(\epsilon) \Rightarrow |f(z)| < \epsilon.$$
Picking $\epsilon =1$ we have that: $|f(z)| < 1$ when $|z|>N(1)$.
2) Using the fact that $f$ is continuous everywhere it is naturally bounded on the disc $\overline {B(0,N(1))}$.
Using 1) and 2) we have that $f$ is bounded on $\mathbb C$, and we may apply Louville's theorem.