The $8$ Sylow subgroups of order $7$ each contain $6$ elements of order $7$ together with the identity. Since an element of order $7$ generates a group of order $7$ these elements are all distinct. There are $8 \times 6=48$ elements of order $7$.
You use the fact that the group is simple in asserting that there are $8$ subgroups of order $7$. It is always possible to have $1$ subgroup of a given order (so long as it is a factor of the group order e.g. the cyclic group of order $168$) - but a single subgroup of order $7$ would be normal, and the group would not be simple.
Let's start by finding one subgroup of order $8$. Since this will have the correct order, it will be a Sylow $2$-subgroup. Note that $S_5$ is the group of permutations of $\{1,2,3,4,5\}$. If we view $\{1,2,3,4\}$ as the vertices of a square, and consider the dihedral group of that square, then we have identified a subgroup of order $8$.
Since all Sylow subgroups of a given order are conjugate, this means that all Sylow $2$-subgroups of $S_5$ are isomorphic to $D_8$, the dihedral group of a square. The others are found by choosing different subsets of $\{1,2,3,4,5\}$ for the vertices of the square. Also note that reordering the vertices as, for example, $1,2,4,3$ versus $1,2,3,4$ gives us a different subgroup.
There are $\displaystyle {5 \choose 4} = 5$ ways to choose $4$ vertices from $5$ elements. For each choice of $4$ vertices, there are $6$ ways to order them (assuming we view cyclic shifts such as $1,2,3,4$ and $2,3,4,1$ as the same ordering), resulting in $3$ distinct dihedral groups (because "opposite" pairs of orderings such as $1,2,3,4$ and $1,4,3,2$ result in the same group; just flip the square upside-down to go from one ordering to the other). Thus there are $5 \cdot 3 = 15$ distinct subgroups of order $8$.
Best Answer
So a Sylow 7-subgroup of $|S_7|$ is going to have order 7, by the prime decomposition you write down in your question.
Any group of prime order is cyclic, so all the Sylow subgroups are cyclic. Conversely, if you have a subgroup of order 7 then it's always going to be Sylow.
Therefore, you just need to identify the subgroups that are isomorphic to $C_7$. For this it suffices to identify all the elements of order 7, that is, the 7-cycles.
There are $7!/7 = 6!$ of these, and since each copy of $C_7$ has exactly one element not of order 7 (and no element of order 7 appears in two different copies of $C_7$) there are $6!/6 = 5!$ subgroups of order 7.