Am I the only one constantly forgetting the Eisenstein criterion?

It's memorable if you view the proof as a consequence of the unique factorization of prime products.

The first part of the hypothesis says: $\bmod p\!:\ f \equiv a x^n\,$ is (an associate of) a prime power $x^n$

By uniqueness a $\rm\color{#c00}{proper}$ factorization has form $\, gh\equiv (b x^i) (c x^j)\,$ for $\,\color{#c00}{i,j \ge 1}$

But $\,\color{#c00}{i,j \ge 1}\,\Rightarrow\, p\mid g(0),h(0)\,\Rightarrow\, p^2\mid g(0)h(0)\!=\!f(0),\,$ contra hypothesis.

Said in ideal language it is $\,(p,x)^2\equiv (p^2)\,\pmod{\! x}$

Remark $ $ This view immediately leads to the discriminant-based test for finding shifts $x\mapsto x+c$ that are Eisenstein, e.g. see this answer.

The way I always remember it is via the Newton polygon.

The basic idea is this: You plot a graph of dots with coordinates $(n,v_p(a_n))$ for $a_n$ the coefficients of your polynomial and $v_p$ the valuation with respect to $p$. Then you look at the lower convex hull, if this is a single line segment with slope $-1/n$ then the polynomial is irreducible, http://www-users.math.umn.edu/~garrett/m/number_theory/newton_polygon.pdf for more details.

Now this probably sounds more complicated! But its very geometric and thus easier to visualise whats going on (it also generalises quite well!) and just remembering what the picture has to look like is more than enough to recover the condition.