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This question is related to Theorem $1$ in
my old answer.

All primes $p=9m^2+3m+1$ with $m\in \mathbb{Z}$ up to $1000$ can be found in this paper (see Table $1$ on the page $7$).

davidoff303
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    It seems there are infinitely many such prime numbers. – Michael Rozenberg May 28 '19 at 12:31
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    If there were a theory about when a prime is congruent to a sum of quadratic residues it might help. –  May 28 '19 at 17:58
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    The only non-trivial result in this context , apart from Dirichlet's theorem that every arithmetic progression $an+b$ with coprime $a$ and $b$ produces infinite many primes, is that there are infinite many primes of the form $a^2+b^4$, but this is a polynomial with two unknowns. – Peter May 30 '19 at 08:19

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The Bunyakovsky conjecture implies that there are infinite many primes of this form, but the only solved case is degree $1$ (Dirichlet's theorem).

For no integer polynomial with degree $d>1$, it is known that it produces infinite many primes.

Peter
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