Let $p|2^{17} - 1$, then we have $2^{17} \equiv 1 \pmod p$, but we also have from Fermat's Little Theorem $2^{p-1} \equiv 1 \pmod p$.
From Lagrange Theorem for group order we get that $2^{\gcd(p-1,17)} \equiv 1 \pmod p$.
Obivously the greatest common divisor is $17$, because otherwise (if the gcd were $1$) then $2 \equiv 1 \pmod p$, which is impossible.
Then we have $17 \mid p-1$ and because $p-1$ is even we have $34\mid p-1$. So we need to check all prime numbers of the form $p=34k + 1 \quad \forall k \in \mathbb{N}$ such that $p \in (1,\sqrt{2^{17} - 1}]$. Which reduces the number of divisors to check. Actually there are just $4$ primes in this inteval, those are $103, 137, 239, 307$. You can check that none of them divides $2^{17} - 1$, implying that $2^{17} - 1$ is prime.
Also you can check how Euler proved that $2^{31} - 1$ is prime, and apply the same method on $2^{17} - 1$.
Actualy $2^{17} - 1 = 131071$ isn't that big number, even without knowledge in number theory it isn't so much hord work to do.
And at last as other users have suggested use Lucas-Lehmer Test.
Full proof that there is only one solution $m=1, p=3, n=6$. $2^n\mod 7$ is $1,2,4,1...$ as $n=0,1,2,3,...$, $m^3\mod 7$ is $0,1,-1$. So $m^3\mod 7 =2^n\mod 7$ implies $n=3k$, $m\equiv 1, 2, 4\mod 7$. So you have $7p^2=2^{3k}-m^3=(2^k-m)(2^{2k}+2^km+m^2)$. Since $m^3<2^{3k}$, $m<2^k$. So $2^k-m$ is a positive divisor of $7p^2$, whence $2^k-m=1,7,p, 7p, p^2$ or $7p^2$, and, resp. $2^{2k}+2^km+m^2=7p^2, p^2, 7p, p, 7$ or $1$. The last option is not possible. If $2^k=m+1$, then $(m+1)^2+(m+1)m+m^2=3m^2+3m+1=7p^2$, so $m\equiv 1\mod 7$, $m=7s+1$ ($m\equiv 2, 4$ are impossible since $m$ is odd). Hence
$147s^2+63s+7=7p^2$ or $21s^2+9s+1=p^2$. So $(p-1)(p+1)=3s(7s+3)$. Or $=3(2^k-2)/7(2^k+1)$. Note that $(p-1)(p+1)$ is divisible by $4$. The only way $3(2^k-2)/7(2^k+1)$ is divisible by $4$ is $k=1$. Then $m=1$ which gives $7p^2=7, p=1$, impossible.
If $2^k-m=p$, then $(m+p)^2+(m+p)m+m^2=3m^2+3mp+p^2=7p$. So $p<7$ and you can check $p=2,3,5$.
If $p=2$, $7p^2=28$, $2^{k}-m=2$, $3m^2+6m+4=14$, no integer solution. If $p=3$, then $3m^2+9m+9=21$, $m^2+3m-4=0$, $m=-4, 1$. Only $m=1$ works, $k=2$. So $n=6$. Check $2^6-(1)^3=63=9*7=3^2*7$. So $m=1, n=6$ works.
Finally, $p=5$, then $3m^2+15m+25=35$, $3m^2+15m-10=0$, no solution.
Similarly consider the cases $2^k-m=7, 7p, p^2$. For example if $2^k=m+7p$ then $(m+7p)^2+m(m+7p)+m^2=p$ which is impossible. Same for $2^k-m=p^2$.
The case $2^k-m=7$ gives
$(m+7)^2+(m+7)m+m^2=p^2$, $3m^2+21m+49=p^2$. Modulo $7$,
$m^2=0,1,4,2$, then the LHS $\equiv 0, 5, -1$ which can be $=p^2\equiv 0,4,2$ only if $p\equiv 0 \mod 7$, so $p=7$ being a prime. Hence $m=0$ or $m=-7$ but that contradicts the assumption that $m+7$ is a power of $2$.
Best Answer
Hint. We have that $f(2)=31$ and $f(3)=121=11^2$. For any integer $n>3$, show that $$\left(n^2 + \left\lfloor\frac {n}{2}\right\rfloor\right)^2 < f(n)=\frac {n^5 - 1}{n - 1} < \left(n^2 + \left\lfloor\frac {n}{2}\right\rfloor+1\right)^2.$$