Assume there are only finitely many primes $\equiv -1 \mod 5$ and let $P$ be their product. Let $q$ be any prime divisor of $(2P^2)^2-5$. Then $1= \left( \frac{5}{q} \right) = \left( \frac{q}{5} \right)$. Thus $ q \equiv \pm 1 \mod 5$. Additionally $(2P^2)^2-5 \equiv -1 \mod 5$, so we can conclude that at least one times the $q \equiv -1 \mod 5$ case happens when $q$ runs through all prime divisors of $(2P^2)^2-5$. But $q$ is trivially coprime to $(2P^2)^2-5$, a contradiction to the assumption. And primes $\equiv -1 \mod 5$ always end in $9$.