Let's look at the gcd of $5p$ and $x$. If $GCD(5p,x)=1$ we have $5p+x|(5p)^n+x^n-5^{n-1}x^n$ thus we have $5p+x|5^{n-1}-1$ so $5p+x|4$ but this is impossible. Now we check the cases when $GCD \in \{ 5,p,5p \}$. $1)$ $x=5\cdot k$. Inserting back we have $p+k|p^n+5^{n-1}k^n$. We look at the $GCD$ of $p$ and $k$. If it is $1$ we similarly get $p+k|4$ so we get $p=2,3$. Checking these we can verify the solution: $(p,x)=(2,10)$. If not, we check again the cases... $2)$ $x=p \cdot k$. Inserting back ... got bored here.