Here is what I think you are asking: for each natural number $m$, are there infinitely many primes $p \equiv 1 \bmod 6m$ such that $60m^2 + 6m - 1 \bmod p$ is a quadratic residue?

To avoid being distracted by the algebraic expressions, set $a = 60m^2 + 6m-1$ and $b = 6m$. I think you are asking if there are infinitely many primes $p \equiv 1 \bmod b$ such that $a \bmod p$ is a quadratic residue. I'll show this can be done for arbitrary nonzero integers $a$ and $b$.

For each nonzero integer $n$, quadratic reciprocity implies $n \bmod p$ is a quadratic residue if (not only if) $p \equiv 1 \bmod 4|n|$. So it suffices to find infinitely many primes $p$ such that
$$
p \equiv 1 \bmod b, \ \ \ p \equiv 1 \bmod 4|a|.
$$
These congruences both hold for a prime $p$ such that
$$
p \equiv 1 \bmod 4|a|b.
$$
Dirichlet's theorem tells us there are infinitely many primes $p$ satisfying that last congruence condition, and for all them you'll have
$p \equiv 1 \bmod b$ and $a \bmod p$ is a quadratic residue.

Actually, it is overkill to appeal to Dirichlet's theorem here, because we are seeking primes satisfying a congruence condition of the form $p \equiv 1 \bmod N$, and in that case the infinitude of such primes follows purely algebraically using values of the $N$th cyclotomic polynomial by modifying Euclid's proof of the infinitude of the primes.