Primes $p=n^6+1$

Question

Which is the least odd prime $p=n^6+1$ for some $n\in\mathbb N$? I have tested for $n\leq 10,000$ without finding any.

Due to a conjecture of Bunyakovsky there are an infinite number of such primes, since $X^6+1$ is irreducible over $\mathbb Z$.

Answer 1

If $n>1$ $$n^6+1 = \underbrace{(n^2+1)}_{>1}\underbrace{(n^4-n^2+1)}_{>1}$$ then $p$ is not a prime.