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 $x^2=c$,
$$(ra+s)(ta+u) = (ru+st)a + (su+rtc).$$
If every element in the field is of the form $ax^2+bx+c + <f(x)>$ and there are three choices for $a$, three choices for $b$, and three choices for $c$, and different choices give rise to different elements, then by some very elementary combinatorics there are $3 \times 3 \times 3=27$ possible choices.
Best Answer
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!