If you don't know about ramification or valuations and things, there is still a great easy way to "see" Eisenstein from the $\mathbb{Z}[x]$ case. This is presented in Integers, Polynomials, and Rings by Ron Irving as a series of exercises to get the student to guess Eisenstein on their own.
First examine the case $x^n-p$ where $p$ is a prime. If this factors, then there are $f(x)=\sum_{k=0}^m a_kx^k$ and $g(x)=\sum_{k=0}^r b_kx^k$ for which $f(x)g(x)=x^n-p$. First, you get that $a_0b_0=p$ and hence without loss of generality $a_0=p$ and $p$ does not divide $b_0$ (we'll merely use this fact rather than $b_0=1$).
Now since $a_0b_1+a_1b_0=0$ we get that $p|(a_0b_1+a_1b_0)$ and $p|a_0$ so $p|a_1b_0$. This means $p|a_1$ or $p|b_0$, but $p$ does not divide $b_0$, so $p|a_1$. You can keep bootstrapping this argument up all the coefficients $a_k$ to get that $p|a_m$ the top coefficient, which is a contradiction since $a_mb_r=1$.
But what was really used here? Not much. So you could repeat the exact same argument with $x^n-pm$ where $(p,m)=1$. But that still isn't as general as you could go. You didn't need that $\sum_{j+k=l} a_jb_k=0$, you only needed that $p$ divided that sum to make the argument work, which is the same as checking that the polynomial $x^n+c_{n-1}x^{n-1}+\cdots +c_0$ has all $c_i$ divisible by $p$ (and $c_0$ not divisible by $p^2$).
Yet again, this still isn't as general as the argument allows you to go, because the contradiction only came from the fact that $p$ did not divide the leading coefficient, it wasn't that it was $1$.
So merely extrapolating what made the proof that $x^n-p$ is irreducible in $\mathbb{Z}[x]$ work gives you the full Eisenstein in $\mathbb{Z}[x]$. Lastly, if you want to go to $R[x]$, you need to figure out which facts about division you needed about $\mathbb{Z}$. But Eisenstein really is a criterion invented artificially in $\mathbb{Z}$ that could be guessed from examining what the key properties of one proof were.