Show $x-1$ is irreducible in $\mathbb{Z}_8[x]$

Aha, together we have the necessary pieces! You've remarked (by looking modulo $2$) that any factorization of $x-1$ in $\Bbb Z_8[x]$ must look like $$ x-1 = (1-x+2f(x))(1+2g(x)). $$ However, $1+2g(x)$ is a unit, since its reciprocal is $1-2g(x)+4g(x)^2$. Therefore $x-1$ is indeed irreducible in $\Bbb Z_8[x]$.

First of all I have to admit that I don't know what's an irreducible polynomial over a non-integral domain.
But if I understand well the OP wants to show that $X-1$ can't be written as a product of two non-invertible polynomials (of degree at least one).
Suppose the contrary, and write $X-1=fg$ with $\deg f\ge1$, $\deg g\ge1$.
If we work over a ring $R$ having only one prime ideal $\mathfrak p$ (as it is $\mathbb Z/8\mathbb Z$), then we arrive to a contradiction immediately: in $R/\mathfrak p$ we have $X-\overline 1=\overline f\overline g$ hence $\deg\overline f=0$ and $\deg\overline g=1$ (or vice versa). Then all coefficients of $f$, excepting $f_0$, belong to $\mathfrak p$, so they are nilpotent. But obviously $f_0$ is invertible, so $f$ is invertible.

Remark. We showed that $X-1$ is "irreducible" over any commutative ring having only one prime ideal (in particular for $\mathbb Z/p^n\mathbb Z$ with $p$ prime, $n\ge1$). One can't generalize this to commutative rings with at least two prime ideals: in $(\mathbb Z/6\mathbb Z)[X]$ we can write $X-1=(3X-1)(2X+1)$.