Let $f(x)=a_0 +a_1x+\dots +a_nx^m \in\mathbb Z[x]$.
Suppose $\exists p$ prime: $p\mid a_1$, $p\mid a_2$, $\dots$, $p\mid a_n$, but $p^2 \nmid a_n$ and $p\nmid a_0$.
Prove $f(x)$ is irreducible in $\mathbb{Q}[x]$.
Note that this is not Eisentein's criterion, but it is very close.
"Proof" (by contradiction)
We assume $f(x)=a_0 +a_1x+\dots +a_nx^m \in\mathbb Z[x]$, $\exists p$ prime: $p\mid a_1$, $p\mid a_2$, $\dots$, $p\mid a_n$, but $p^2 \nmid a_n$ and $p\nmid a_0$, and $f(x)$ is reducible in $\mathbb{Q}[x]$.
if $f(x) $ is red. over $\mathbb{Q}$ we know it will be red in $\mathbb{Z}[x]$ so $\exists g(x),h(x)\in\mathbb Z[x]$ s.t. $$ f(x)=g(x)b(x) \text{ where } 1 \leq \deg g(x),\deg h(x) <n$$
Say $g(x)=b_rx^n+\dots+ b_0$ and $h(x)=c_sx^s+\dots +c_0$ now, $$ p\nmid a_0 \equiv \neg (p\mid c_0 t_0)\equiv \neg (p\mid c_0 \vee p\mid t_0)\equiv p\nmid c_0 \wedge p\nmid t_0 $$
$$p^2 \nmid a_n\equiv p^2 \nmid c_s t_r \equiv p^2 \nmid c_s \wedge p^2 \nmid c_r$$
also, $p\mid a_n$ so $p\neq 0,\pm1 $ so $p\mid c_r r_s $
$\vdots$
Contradiction somewhere
$\vdots$
Expanding my comment into an answer
Let $g(x) = x^mf(x^{-1}) = a_0 x^m+\cdots+ a_n$ be the reciprocal polynomial of $f(x)$. Note that this is a polynomial with coefficients in reverse order of $f(x)$. And in this particular case $g(x)$ is irreducible by Eisenstein criterion.
We claim: If reciprocal polynomial of $f(x)$ is irreducible then $f(x)$ is irreducible.
Proof: For the sake of contradiction assume $f(x)$ is irreducible. Let $g(x)= a(x) b(x)$ with $\deg a(x)= i$ and $\deg b(x)= m - i, 0< i < n$. Then $g(x)= x^m a(x^{-1}) b(x^{-1})$. So $f(x^{-1}) = a(x^{-1})b(x^{-1)}$. And since $x$ is treated as a formal symbol that will imply $f(x)$ is reducible. Which finishes the proof.
Though if you don't want to use Eisenstein your strategy might work, I will think about it.