I know that checking $f = x^4+x^3+x^2+x+1 \in \mathbb{Q}_3[x]$ over its residue field $\mathbb{F}_3$ is not enough since it has degree $4$. I also know that $f$ is supposed to be a polynomial where a primitive $5$-th root of unity vanishes but I would like to show the irreducibility more directly. Stuff like the Eisenstein criterion cannot be applied to.
Could you please help me with this problem?
Here’s how you can do it, and I think it’s the best way.
First, because $\Bbb Z_3$, the $3$-adic integers, are a Unique Factorization Domain, irreducibility over $\Bbb Q_3$ is the same as irreducibility over $\Bbb Z_3$. (I’ve left off some details, but it’s the same as the story of $\Bbb Q$ versus $\Bbb Z$.)
Second, what you say about irreducibility over $\Bbb F_3$ not being enough to conclude irreducibility over $\Bbb Z_3$, that’s wrong. A factorization over $\Bbb Z_3$ will induce a factorization over $\Bbb F_3$.
Now I show why $x^4+x^3+x^2+x+1$, which you know is $(x^5-1)/(x-1)$, is $\Bbb F_3$-irreducible. Its roots are fifth roots of unity unequal to $1$. And it’s irreducible if and only if adjoining a fifth root of unity requires a degree-four extension of the base field $\Bbb F_3$. How do you decide this? See what the first field $\Bbb F_{3^m}$ is that contains fifth roots of unity.
How do you do that? Look at $(\Bbb F_{3^m})^\times$, the (cyclic) group of nonzero elements, $3^m-1$ in number, and see whether it has an element of period five, i.e. see whether $5|(3^m-1)$, i.e. see whether $3^m\equiv1\pmod5$. Oh of course, that’s $81=3^4$. (You could also have asked for the period of $3$ in $(\Bbb F_5)^\times$.) No matter what, there’s your proof.