I am having a huge issue that I can't fix in my argument. I will present my argument first then I will mention why I am having troubles fixing it.
Let $p$ be a sylow $p$-subgroup of $G$. Then $n_p \mid (p + 1)$ and $n_p = 1 \pmod p$. If $n_p = 1$, then $p$ is normal and we are done. Otherwise we have $n_p = kp + 1$. Since $(kp + 1) \mid (p + 1)$ so this imply that $n_p = p + 1$.
Therefore there are $p + 1$ sylow $p$ each having size $p$, so that means each distinct $p$ subgroups have intersection $\{e\}$. Therefore the union of all $p$ subgroups gives $(p + 1)(p - 1) + 1 = p^2$ elements. Therefore this leave $|G| - p^2 = p$ elements remaining outside all $p$-sylows.
Let $A$ be the set of elements that don't lie in any $p$-subgroup of $G$. $|A| = p$. Let $q$ be any prime that divides $p + 1$. Then A must contain some element $g$ of order $q$. Since $n_p = p + 1$, so this means that $N_G(P) = P$, therefore $g$ isn't in $N_G(P)$.
Let $x$ be a generator of $P$. Then all powers $x^i$ for $i \in \{1,\ldots,p - 1\}$ are generators of $P$. Since $g$ isn't in $N_G(P)$ so none of the $x^{i}$ centralizes $g$. Therefore $\{g,\ldots,x^{p - 1}gx^{-(p - 1)}\}$ are all distinict. All of those elements have order $q$. Since each of them must lie in $A$, so $A = \{g,\ldots,x^{p - 1}gx^{-(p - 1)}\}$. I can't really now deduce that $p + 1$ is a power of $q$ (for example consider $|G| = (5)(6)$). Any suggestions would be nice.
Note that conjugation by $x$ is an automorphism, so the order of $g$ is the same as the order of $xgx^{-1}$, which in turn is the same as the order of $x^2gx^{-2}$, and so forth. Since you have shown that $A = \{g, xgx^{-1}, \dots , x^{p-1}gx^{-(p-1)}\}$, this shows that every element of $A$ has the same order.
If we denote this order by $a$, then the elements of $G$ have orders only $1,p,a$. This implies that $a$ is a prime distinct from $p$, since if $a$ was composite then there would also be elements with order equal to each divisor of $a$. ($a$ cannot be equal to $p^2$ for obvious reasons)
This in turn implies that $p+1 = a^k$ is a power of this prime, since since for any prime divisor of $|G|$ there is always an element of that order. We must conclude that $A \cup \{1\}$ is the unique Sylow $a$-subgroup for $G$, and is thus normal of order $p+1$.