Lemma 1
If $F$ is a field and if $p$ is an irreducible polynomial with coefficients in $F$, then $F[x]/(p)$ is a field.
Proof
A polynomial $p\in F[x]$ is irreducible if and only if $(p)$ is a maximal ideal in $F[x]$. A quotient $R/I$ of a commutative ring $R$ by an ideal $I$ is a field if and only if $I$ is a maximal ideal in $R$. Q.E.D.
Lemma 2
In the context of Lemma 1, $F\subseteq F[x]/(p)$ so $F[x]/(p)$ is a vector space over $F$. The dimension of this $F$-vector space is $\text{deg}(p)$.
Proof
Firstly, technically $F$ is not a subset of $F[x]/(p)$. However, the canonical homomorphism $F[x]\to F[x]/(p)$ restricts to a homomorphism of $F\to F[x]/(p)$ (since $F$ is a subset of $F[x]/(p)$). Furthermore, this map $F\to F[x]/(p)$ is injective (an element of $F[x]$ is sent to $0$ under $F[x]\to F[x]/(p)$ if and only if its divisible by $p$; clearly, no constant polynomial is divisible by $p$ except $0$). Therefore, we can view $F$ as a subset (in fact, a subfield) of $F[x]/(p)$.
We claim that the set ${\cal B}=\{1+(p),x+(p),x^2+(p),\dots,x^{\text{deg}(p)-1}+(p)\}$ is a basis for $F[x]/(p)$ as an $F$-vector space. Firstly, if $q+(p)\in F[x]/(p)$ (for $q\in F[x]$), then the division algorithm implies that $q=pa+b$ for $a,b\in F[x]$ and $b=0$ or $\text{deg}(b)<\text{deg}(p)$. Since $q+(p)=b+(p)$ and $b+(p)$ is clearly in the span of ${\cal B}$, it follows that ${\cal B}$ spans $F[x]/(p)$ as an $F$-vector space.
An equation of linear dependence for ${\cal B}$ is equivalent to a polynomial $q$ of degree less than that of $p$ equalling zero in $F[x]/(p)$. If the coefficients of $q$ were non-zero, then this would be a contradiction as $p$ cannot divide any non-zero polynomial of smaller degree. Therefore, $q=0$ and ${\cal B}$ is linearly independent.
Therefore, $F[x]/(p)$ is $\text{deg}(p)$-dimensional as an $F$-vector space. Q.E.D.
Lemma 3
If $V$ is a vector space of dimension $n$ over a finite field $F$ with $k$ elements, then $V$ has $k^n$ elements.
Proof
Exercise! Q.E.D.
Exercise: What is the number of elements in $\mathbb{Z}_2[x]/(x^3+x^2+1)$?
I hope this helps!
Best Answer
The elements you list need not be all distinct, so your list need not be the full list of elements of your field.
For example, consider $\mathbf{F}_7$; the squares are $1$, $2$, and $4$. In particular, $-1$ is not a square, so you can take $c=-1$; then your list of elements consists only of $0$, $a$, $-1$, $-a$, and $1$, which is only $5$ elements, not the required $49$.
In fact, your list can never be all the elements, because $c^p = c$ holds, so you are repeating a lot of elements, and not getting all of them.
(Added. The best you can hope for is if $c$ is a primitive element of $\mathbf{F}_7$; that is, it generates the group of units. You still get only $2p-1$ of the required $p^2$ elements, though.)
Instead, you want to think of the elements as being the result of evaluating polynomials at $a$; since every polynomial can be written as a multiple of $x^2-c$ plust a remainder, the elements of the field will be of the form $ra+s$, with $r,s\in\mathbb{F}_p$. You add them the usual way, $(ra+s) + (ta+u) = (r+t)a+(s+u)$, with addition on the right being addition modulo $p$, and you multiply them by using the fact that $x^2=c$, $$(ra+s)(ta+u) = (ru+st)a + (su+rtc).$$