Theorem: Every finite field has order $p^n$.
Proof:
Assume by contradiction a field $F$ of order $k$ such that $k$ has two distinct primes (of orders $m$ and $n$ dividing it), namely $p$ and $q$. Since $<F, +>$ is an (abelian) group of order k, we have a Sylow $p$-subgroup of order $p^m$. By Cauchy's theorem (existence of an element of prime order if it divides the group order) we have an element of order $p$ in the subgroup, hence in the group. i.e., $a +_p a = 0$ (where $+_p$ is addition to itself p times). The same logic with another distinct prime gives $b +_q b = 0$. WLOG take $p < q$
Consider $c \in F$ such that $c*a=b$ (field multiplication). This exists because $F/\{0\}$ possesses group structure. Then do $c(a +_p a) = c*0=0$ since the zero element of a field "absorbs multiplicatively". But then the LHS becomes $(b +_p b) = 0$ by distributivity and the construction of the element $c$. This implies that $b$ is of order $p$, which it cannot be because it was constructed to be within a Sylow $q$-subgroup and $p$, being different from $q$, can divide no power of $q$, so $b$ being of order $p$ contradicts Cauchy's theorem.
Therein lies the problem (as I see it) with a field with order not a prime power.
The proof looks right, but I don't understand why you felt the need of using Sylow's theorems. Why did you not apply Cauchy's theorem directly to $(F,+)$?
There are some problems concerning your notation. You should have written $F\setminus\{0\}$ instead of $F/\{0\}$. And I don't think that there is a need of introducing the $+_p$ notation. You could have just written $pa$ instead $a+_pa$.