= Solution
Suppose there were only finitely many <prime numbers> congruent to $2$ modulo $3$, and list them as $p_1,\ldots,p_k$. Consider
$$
N=3p_1\cdots p_k-1.
$$
No $p_i$ divides $N$, and $3$ does not divide $N$. In the <prime factorization> of $N$, not every prime factor can be congruent to $1$ modulo $3$, since their product would then also be congruent to $1$, whereas
$$
N\equiv2\pmod3.
$$
Thus some prime factor is congruent to $2$ modulo $3$, contradicting the completeness of the list. Hence
$$
\boxed{\text{there are infinitely many primes of the form }3n+2}.
$$
Now let $p$ be prime. If $p\ne3$, then either $p=2$, giving
$$
2p^2+1=9,
$$
or $p$ is not divisible by $3$, so $p^2\equiv1\pmod3$. In the latter case
$$
2p^2+1\equiv0\pmod3
$$
and is greater than $3$, hence is composite. For $p=3$ the number is $19$, which is prime. Therefore
$$
\boxed{p=3\text{ is the only possibility}}.
$$
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