The only finite fields that equal the rings $\Bbb Z/n\Bbb Z$ are the ones where $n$ is a prime. Because if $n=ab$ then $\overline{a}\cdot\overline{b}=\overline{0}$ in $\Bbb Z/n\Bbb Z$ and we know that can't happen in a field. On the other hand if you adjoin a root of an irreducible polynomial of degree $2$ to $\Bbb Z/3\Bbb Z$ then you get a degree two extension of $\Bbb Z/3\Bbb Z$ which has nine elements, but is not equal to $\Bbb Z/9\Bbb Z$.
There are two ways you might want to represent a field of order $q^k$ (where $q\in \mathbb{N}$ is a prime and $k \gt 0$ a positive integer).
One is to imagine a simple algebraic field extension of $\mathbb{Z}_q$ which has a basis $\{1,\alpha,\ldots,\alpha^{k-1}\}$, namely $\mathbb{Z}_q[\alpha]$ where $\alpha$ satisfies an irreducible monic polynomial of degree $k$ over $\mathbb{Z}_q$:
$$ \alpha^k + c_1 \alpha^{k-1} + \ldots + c_{k-1}\alpha + c_k = 0 $$
Any element of this field extension can be represented in terms of the basis, and the addition is just the same as the addition of the $k$-dimensional vector space over $\mathbb{Z}_q$. But when we multiply, we have to perform substitutions to eliminate powers $\alpha^k$ and higher by using:
$$ \alpha^k = - c_1 \alpha^{k-1} - \ldots - c_{k-1}\alpha - c_k $$
The other way to think about it is to start with an irreducible polynomial $p(x)$ of degree $k$ over $\mathbb{Z}_q$ and construct $\mathbb{Z}_q[x]/p(x)$ as a quotient ring. Since the ideal generated by $p(x)$ is maximal, the quotient ring is a field and has dimension $k$ over $\mathbb{Z}_q$ as a vector space.
These two constructions are equivalent, with $\alpha$ being a root of:
$$ p(x) = x^k + c_1 x^{k-1} + \ldots + c_{k-1}x + c_k $$
and identified with $x \bmod{p(x)}$ in $\mathbb{Z}_q[x]/p(x)$.
It turns out that all field extensions of degree $k$ over $\mathbb{Z}_q$ are isomorphic, so as a computational convenience the irreducible polynomial $p(x)$ may be chosen to be simple in some way (e.g. having as few nonzero coefficients as possible).
It's not hard to come up with a monic irreducible polynomial of degree $3$ over $\mathbb{Z}_2$ in order to construct a field of $8$ elements. You only need to check for divisibility by linear (first degree) factors to be sure of irreducibility.
Best Answer
I don't think there is any general procedure to find an irreducible polynomial of degree $n$ over $\mathbb{Z}_p$. However, any such polynomial $p(x)$ works, i.e. it will produce a field $\mathbb{Z}[x]/(p(x))$ of order $p^n$.