Assume the contrary that for some $p\ne 5$, if $q\mid p^5-1$ is a prime then $5\nmid q-1$. Observe that $q\ne p$ and by Fermat, $p^{q-1}\equiv 1\pmod{q}$, so $p^{(5,q-1)}= p \equiv 1\pmod{q}$, as such $q\mid p-1$. Next, \[ p^5 - 1 = (p-1)\underbrace{(p^4+\cdots+p+1)}_{:=A}. \]Note that ${\rm gcd}(p-1,A)\mid 5$. So, it suffices to verify that $A$ has a prime divisor other than $5$, which would yield a contradiction---since such a divisor . If $v_5(p-1)\ge 2$, then $p^5+\cdots+p+1\equiv 5\pmod{25}$, which immediately yields the existence of such a prime. Otherwise, if $5\mid \mid p-1$ then $v_5(p^5-1)=2$ by LTE, so $5\mid \mid A$ and since $A>5$, we are done.

Why $p\neq 5$? Even for $5^5 - 1 = 2^2 \cdot 11 \cdot 71$ things seem fine. And I think I never used that $p$ is prime (or least at most that $5$ does not divide $p$). Anyway, for the problem, suppose not and let $q$ be a prime divisor of $p^5 - 1$. By Fermat we have $p^{q-1} \equiv 1$, so the order of $p$ mod $q$ divides both $5$ and $q-1$ and hence must be equal to $1$ (as $5$ does not divide $q-1$ by our assumption). In particular, $q$ divides $p-1$ and so all prime divisors of $p^5 - 1 = (p-1)(p^4 + p^3 + p^2 + p + 1)$ must divide $p-1$. Note that the greatest common divisor of both multipliers is $5$ (if $p\equiv 1 \pmod d$, then $p^4 + p^3 + p^2 + p + 1 \equiv 5 \pmod d$) and so no prime divisors of $p^5-1$ can divide the second multiplier apart from possibly the integer $5$. Hence this forces to have $p^4 + p^3 + p^2 + p + 1 = 5^k$ for some integer $k \geq 2$ (as $p\geq 2$). However, the left-hand side is never divisible by $25$, contradiction. (Indeed, mod 5 requires $p=5m+1$ for some $m$ and then in $p^4 + p^3 + p^2 + p + 1 = 5 + 50 m + 250 m^2 + 625 m^3 + 625 m^4$ all summands apart from the first one are divisible by $25$.)

Just use zsigmondy there is a prime with order 5!!!!

There is a prime divisor $q$ of $p^5 -1$ which does not divide $p-1$ due to Zsigmondy then $5 |q-1$ because $q | p^{q-1}-1$(FLT) so we are done.

By Zsigmondy's, there exists a primitive prime factor $q$ of $p^5 - 1$. $p^5 \equiv 1\mod{q} \implies \text{ord}_q(p) \mid 5$. Since $\text{ord}_q(p) \neq 1$ (as $q \nmid p - 1$), $\text{ord}_q(p) = 5 \implies 5 \mid q - 1$ by Fermat's Little Theorem, which means $q \equiv 1\mod{5}$, which implies the result.