Premise: not a native speaker.
So: I really don't know how to find all the polynomials except to calculate them all:
\begin{equation}x^{2}+ax+b \end{equation} \begin{equation} a,b ∈ {(0, 1, 2, 3)} \end{equation}
\begin{equation} \begin{split} 1)x^{2}\\ 2)x^{2} + x\\ 3)x^{2} + 1\\ 4)x^{2} + x + 1\\ 5)x^{2} + 2x\\ 6)x^{2} + 2\\ 7)x^{2} + 2x + 1\\ 8)x^{2} + x + 2\\ 9)x^{2} + 2x + 2\\ 10)x^{2} + 3x\\ 11)x^{2} + 3\\ 12)x^{2} + 3x + 3\\ 13)x^{2} + 3x + 1\\ 14)x^{2} + 32 + 2\\ 15)x^{2} + x + 3\\ 16)x^{2} + 2x +3\\ \end{split} \end{equation}
I know that only numbers 3, 4, 6, 9, 12, 13, 15, 16 are Irreducible, because they didn't have roots in $\Bbb{Z}_{4}[x]$.
My question is: is there a more orthodox method to find the irreducible polynomials? Because with larger numbers, this method is not that practical.
It is easier to count all reducible monic polynomials of degree $2$ because they are just the product of two monic linear polynomials, which are $x,x-1,x-2,x-3$ in your cases. So there are ${4 \choose 2} = 6$ products of different linear polynomials. As Robert pointed out, there are also $2$ squares of them. There are $16$ possible monic polynomials in $\Bbb{Z}_{4}[x]$ so there are $8$ irreducible such polynomials. In general, I think this is the efficient way to find all reducible and thus, all irreducible monic polynomials. For polynomials with larger degree, we can start with those with smaller degree first. For example, to know all reducible monic polynomials of degree $3$, we need to know irreducible linear polynomials and for degree $4$, we need to know all irreducible polynomials of degree $1$ and $2$.