Suppose $p>3.$ Then $p$ can be written in the form $6n\pm 1$ for some $n\in\mathbb{Z}^+,$ so (since $a^6\equiv 1\pmod{7}$ for all integer $a$ s.t. $7\nmid a$) \begin{align*}3^p+5^p-1=3^{6n\pm1}+5^{6n\pm1}-1 &\equiv 3^{\pm 1}+5^{\pm 1}-1\pmod{7} \\ &\equiv 3+5-1\text{ or }3^{-1}+5^{-1}-1\equiv 5+3-1\pmod{7} \\ &\equiv 0\pmod{7}\end{align*}i.e. $7\mid 3^p+5^p-1$ for prime $p>3,$ so $3^p+5^p-1$ cannot be prime for those values. Now we only need to check $p=2,3$: $$3^2+5^2-1=9+25-1=33,\quad 3^3+5^3-1=27+125-1=151$$and only the latter of these is prime, so the only satisfactory prime is $\boxed{p=3.}$

Note that there are exactly $\phi ( 7-1) =2$ primitive roots modulo $7$. For example, $3,5$ are distinct primitive roots modulo $7$. Assume that $\gcd( p,6) =1$. Since $3^{p} ,5^{p}$ are distinct primitive roots modulo $7$, $ $we have $3^{p} +5^{p} \equiv \mu ( 7-1) =1\ \pmod 7$ due to the following well-known lemma : Lemma The sum of all primitive roots of modulo $p$ is congruent to $\mu ( p-1)$. Thus we must have $7\mid 3^{p} +5^{p} -1$. So we must have $\gcd( p,6) >1$, which implies $p=2,3$. $3^{2} +5^{2} -1=33$ $3^{3} +5^{3} -1=151$ Hence the answer is $p=3$.

3^p+5^p_1=q P mod6==1,5 then p=6k+5 or 6k+1 If p=6k+1......3^(6k+1)+5^(6k+1)_1 mod7==0 why? Because theorm: a^(p_1) mod(p)==1 then 3^6 mod7==1. And 5^6 mod7==1. 1+1_1=1 If p=6k+5 3^(6k+5)+5^(6k+5)_1 modp==0......why? Because 3^(6k+5)mod7==5 and5^(6k+5)mod7==3 3+5_1 mod7==0 then 3^p+5^p_1 Is a composite number.

Checking $p = 2$ and $p = 3$ we find that $p = 3$ is a solution. Now for $p > 3$, we have: $p \equiv \pm 1 \pmod{6}$ If $p \equiv 1 \pmod{6}$, then by Fermat's little theorem: $3^6 \equiv 1 \pmod{7} \Rightarrow 3^p \equiv 3 \pmod{7}$ $5^6 \equiv 1 \pmod{7} \Rightarrow 5^p \equiv 5 \pmod{7}$ $\Rightarrow 3^p + 5^p - 1 \equiv 3 + 5 - 1 \equiv 0 \pmod{7}$ If $p \equiv -1 \pmod{6}$, then similarly: $3^p \equiv 3^{-1} \equiv 5 \pmod{7}$ $5^p \equiv 5^{-1} \equiv 3 \pmod{7}$ $\Rightarrow 3^p + 5^p - 1 \equiv 5 + 3 - 1 \equiv 0 \pmod{7}$ So, for $p > 3$, the expression is divisible by $7$. Therefore $p = 3$ is the only solution. $\blacksquare$