I am going to take as an assumption that your statement, that when $a_m\neq 0$, all roots have multiplicity $ 1 $, is true. It certainly sounds correct, but I'm no expert on that part.
However, you can easily show the case for $a_m=0$ from that point, as follows:
Suppose that the smallest index that arises is $p^e$. That is,
$$
f(x) = \sum_{i=0}^{m-e} a_ix^{p^{m-i}}
$$
where $a_0=1$ and $a_e\neq 0$. Now, let $y=x^{p^e}$. This allows us to write
$$
f(x) = \sum_{i=0}^{m-e} a_i y^{p^{m-i}/p^e} = \sum_{i=0}^{m-e} a_i y^{p^{m-e-i}/p^e}
$$
Let $M=m-e$, so we have
$$
f(x) = \sum_{i=0}^M a_i y^{p^{M-i}} = g(y)
$$
Now, $g(y)$ has, from the assumed statement, $p^M$ roots of multiplicity $ 1 $.
Now is where we bring in the characteristic being $p$, which means that $(x+y)^p = x^p+y^p$. So we seek solutions to $x^p=a$. If $x_1^p=x_2^p=a$, then $x_1^p-x_2^p = (x_1-x_2)^p=0$, and so $x_1=x_2$. So $x^p-a=0$ has one root of multiplicity $p$, and by iterating this process, we find that $x^{p^e}-a=0$ has one root of multiplicity $p^e$.
But our roots of $g(y)=f(x)$ take the form $y=x^{p^e}=a$, so each root of $g(y)$ is associated uniquely with a root of $f(x)$, and has multiplicity $p^e$.
“Since $\mathbb{Z}[x]$ is a PID...”
Ehr... no, it isn’t. $(2,x)$ is not principal.
It’s also not true that $\mathbb{Z}[x]/(q)$ is a field when $q$ is irreducible: $\mathbb{Z}[x]/(x)\cong\mathbb{Z}$ is not a field (this because $\mathbb{Z}[x]$ is not a PID, so $(q)$ need not be maximal). I don’t think your proof can be salvaged at this point.
Note that maximal ideals of $\mathbb{Z}[x]$
have the form $(q(x),p)$, where $q$ is an irreducible polynomial of degree greater than or equal to $1$, and $p$ is a rational prime. The quotient will be a field of positive characteristic.
Your result is true: it follows from Gauss’s Lemma: Gauss’s Lemma says that if $p(x)\in\mathbb{Z}[x]$ is such that the gcd of its coefficients is $1$, and $p(x) = r(x)s(x)$ with $r(x),s(x)\in\mathbb{Q}[x]$, then there exist polynomials $R(x),S(x)\in\mathbb{Z}[x]$, $\deg(R)=\deg(r)$, $\deg(S)=\deg(s)$, such that $p(x) = R(x)S(x)$.
In particular, if a polynomial $p(x)$ is irreducible in $\mathbb{Z}[x]$, then it remains irreducible in $\mathbb{Q}[x]$. In particular, it cannot have rational roots if $\deg(p)\gt 1$.
Also, note that
The Rational Root Theorem is not about irreducible polynomials; it has nothing to say about it, and it just gives you a finite list of possible roots. I don’t think it is related to the result.
Eisenstein’s Criterion allows you to deduce irreducibility in $\mathbb{Q}[x]$: as such, it necessarily implies no roots when the polynomial has degree greater than $1$. So I do not understand why you say its conclusion “does not go as far” as the result at hand. Having no roots in $\mathbb{Q}[x]$ is weaker than being irreducible.
Best Answer
Assume $a$ is a root of polynomial $f(x) \in F[x]$ of multplicity $\ge k$. Then $a$ is also a root of the $f^{(k-1)}$.
Now assume $f(b)=0$ but $f^{(k-1)}(b)\ne0$. Then $f^{(k-1)}$ is not the zero polynomial and $\gcd(f,f^{(k-1)})$ is a nontrivial divisor of $f$.