As pointed out above, 0 is not invertible. However, for any field $K$, $K[x]$ is a PID. Since $f$ is irreducible and $\mathbb{F}_p[x]$ is a PID, $f$ is a prime element of $\mathbb{F}_p$ (if this isn't clear, then sit down and hammer out a simple proof).
The only thing that we need now is the following proposition.
Proposition: In a PID, every nonzero prime ideal is maximal.
Sketch of proof: Suppose that $D$ is a PID and that we have $0\subsetneq P\subseteq M$ for a maximal ideal $M$ of $D$ and a (nonzero) prime ideal $P$ of $D$. Since $D$ is a PID, we have $P=(p)$ and $M=(q)$ for some nonzero $p,q\in M$. Also, since both $M$ and $P$ are prime ideals, it follows that $p$ and $q$ are prime (and hence irreducible) elements of $D$ (if this isn't immediately clear, then sit down and write out a short proof as to why).
However, $(p)\subseteq (q)$ implies that $q \vert p$, implying that $q=up$ for some $u\in U(D)$. Therefore $M=P$ and every nonzero prime ideal of a PID is maximal. $\blacksquare$
So, going back to $f\in \mathbb{F}_p[x]$, we have that $f$ is a prime polynomial in $\mathbb{F}_p[x]$, hence $(f)$ is a maximal ideal. Therefore $\mathbb{F}_p[x]/(f)$ is a field.
Take $p$ a monic non-null polynomial and write its (unique) decomposition as a product of monic irreducible polynomials :
$$p=q_1^{\alpha_1}...q_r^{\alpha_r} $$
Then by the Chinese Remainder theorem :
$$F[x]/(p)\text{ is isomorphic to } F[x]/(q_1^{\alpha_1})\times...\times F[x]/(q_r^{\alpha_r})$$
Now suppose $\alpha_1=...=\alpha_r=1$ (i.e. $p$ is squarefree) then
$$F[x]/(p)\text{ is isomorphic to } F[x]/(q_1)\times...\times F[x]/(q_r)$$
Cannot have non-zero nilpotent elements because it is a product ring of fields : assume $(a_1,...,a_r)^n=0$ in $F[x]/(q_1)\times...\times F[x]/(q_r)$ then $a_i^n=0$ for each $i$ but you are in a field so $a_i=0$.
On the other hand suppose $\alpha_1\geq 2$ then :
$$(q_1,0,...,0)\neq 0 \text{ in } F[x]/(q_1^{\alpha_1})\times...\times F[x]/(q_r^{\alpha_r}) $$
But clearly :
$$(q_1,0,...,0)^{\alpha_1}=(q_1^{\alpha_1},0,...,0)=(0,...,0)\text{ in }F[x]/(q_1^{\alpha_1})\times...\times F[x]/(q_r^{\alpha_r}) $$
So it has a nilpotent element. For a direct proof you can show that the nilpotent elements of a product of rings is the product of the set of nilpotent elements of each ring.
Best Answer
No, I do not think the first one is correct, take $\mathbb{R}[x,y]/(y-x^2)$, that is wherever you see $y$ make it $x^2$, then the quotient becomes $\mathbb{R}[x,x^2]\cong \mathbb{R}[x]$, and not $\mathbb{R}[x]/x^2$ like you said.