Find all integers n such that $(\frac{n^3-1}{5})$ is prime 

Find all integers n such that $(\frac{n^3-1}{5})$ is prime?


My Approach:
I wrote all the prime which i will get after dividing $(n^3-1)$ by $5$.
$n^3-1=10,15,25,35,55,...,215$
which lead me to $n^3=11,16,26,...,216$, then I obtained $n=6$
My doubt is that how to check more value of $n$ without using modular arithmetic because the book I'm referring has not introduced it yet.
Second approach: $\frac{(n^3-1)}{5}=\frac{(n-1)(n^2+n+1)}{5}$
But my second approach too does not lead me anywhere.
This problem is from the book Pathfinder for Olympiad Mathematics

If we split all integer $n$ into $[5m+1,5m+2,5m+3,5m+4,5m+5]$ you can show that only numbers $(5m+1)^3-1$ are divisible by $5$ as $[2^3-1,3^3-1,4^3-1,5^3-1]$ are all not divisible by $5$
Now $$\frac{(5m+1)^3-1}{5}=3m+15m^2+25m^3=m(3+15m+25m^2)$$
and the product $m(3+15m+25m^2)$, can only be a candidate prime if $m=\pm1$
