What is meant by a polynomial that is "irreducible"? And a "prime" polynomial? Let $P(\mathbb{F})$ be the set of all polynomials of one variable over a field $\mathbb{F}$. It is known that such a set, with usual polynomials operations, is a Euclidean domain, that is, a domain that has the Euclidean division algorithm. So we have the following

Theorem. Let $\deg: P(\mathbb{F})\setminus 0 \to\mathbb{N}$ the degree function. Then $(P(\mathbb{F}), \deg)$ is a Euclidean domain, that is:
a) $P(\mathbb{F})$ is a domain;
b)For all $f,g\in P(\mathbb{F}), g\ne 0$, exist unique polynomials $t,r\in P(\mathbb{F})$ such that 
  $$
f(x) = g(x)q(x) + r(x), \text{ with }\deg(r)<\deg(g) \text{ or }r = 0.
$$

What I want to know:
In such a domain, what is meant by a irreducible or prime polynomial? Moreover, if $p, m\in P(\mathbb{F})$ are such that $p\vert m$ and $p$ is irreducible (or prime) and $m$ is monic (coefficient of the higher degree term is $1$), why we have that $\deg\left(\dfrac mp\right)<\deg(m)$, that is, why $p$ can't have zero degree?
 A: I'll talk about the definitions in general domains, and it should be clear how they apply to the polynomial rings you're interested in.
In general, in any domain, $R$, a non-unit element, $p$, is said to be prime if $p\mid ab$ implies $p\mid a$ or $p\mid b$. A non-unit element, $r$, is said to be irreducible if $r=ab$ implies that one of $a$ or $b$ is a unit. 
All primes, $p$, are irreducible. 
Proof: If $p=ab$, then $p \mid ab$, so $p\mid a$ or $p\mid b$, without loss of generality, we may assume $p\mid a$. Then $a=pv$ for some $v\in R$, and $p=pvb$. Since $R$ is a domain, we may cancel to get $1=vb$, so $b$ is a unit. Hence $p$ is irreducible.
On the other hand, it isn't always the case that irreducible elements are prime. This is however a necessary condition for a ring to be a unique factorization domain, and hence it is in fact true in every Euclidean domain. For a proof, see the answers here, one of which gives a direct proof from the Euclidean property.
To sum up the answer to your first question, in a Euclidean domain, an element is prime ($p\mid ab \implies p\mid a\text{ or } p\mid b$) if and only if it is irreducible ($s=ab \implies a$ or $b$ is a unit).
As for your second question, note that polynomials over a field with degree zero are units, and hence aren't considered irreducible. Thus if $p\mid m$, with $p$ prime/irreducible, then $\deg(m/p) = \deg(m) - \deg(p) < \deg m$. (I've assumed you meant $m$ rather than $f$ in the fraction in your question.) 
A: In $P(\mathbb{F})$, an irreducible polynomial $f$ is a a polynomial OF POSITIVE DEGREE that cannot be written as the product of two polynomials of smaller degree. This is the same as being prime (a prime in a ring is an element $p$ such that if $p | ab$ then $p | a$ or $p | b$) and this is a consequence of $P(\mathbb{F})$ being a unique factorization domain, (because it is a Euclidean domain).
The note that Hungerford's Abstract Algebra gives on this matter is:
You could just as well all such a polynomial "prime," but "irreducible" is the customary term with polynomials.
As for your next question, I think you are then asking why an irreducible polynomial cannot be constant since you are assuming that $p$ is irreducible, then asking why it cannot be constant. Irreducible polynomials are nonconstant by definition. The reason is that in $P(\mathbb{F})$, ever constant polynomial (except $0$) is a unit, they have inverses since $\mathbb{F}$ is a field. So, when you talk about a polynomial being irreducible or prime, one runs into the same sort of reason that $1$ is not prime, because the notion of irreducibility is only a useful concept when you ignore the constants since you can always multiple or divide by a constant and nothing really changes much.
I am not sure what exactly is going on with $f$ and $m$, but I hoped I answered your questions.
A: 
$\underline {\text{Irreducible polynomial}} \text{: A non-constant polynomial that can't be factored within a given field or ring}$

so in the integers: $x^2+1$ is irreducible but $x^2-1$ isn't, this is because the latter factors as $(x+1)\cdot (x-1)$ both of which have coefficients that are integer. 
