Inspired by Mehul Arora

$x=2 y^{2} +1$ assuming that $y =1$ (remainder 3)

then $y^{2}=1$ (remainder 3)

$2 y^{2}=2$ (remainder 3)

$x = 2 y^{2} + 1 =3 = 0$ (remainder 3)

meaning x is a multiple of 3

now assuming that $y=2$ (remainder 3)

then $y^{2}=4 = 1$ (remainder 3)

$2 y^{2}=2$ (remainder 3)

$x = 2 y^{2} + 1 =3 = 0$ (remainder 3)

then as long as $y!=0$ (remainder 3) x must be a multiple of 3.

so it must be that either x or y is a multiple of 3; but the question asks for only prime numbers:

either $x=3$ and according to the equation, $y=1, -1$ which are not primes

or $y=3$ so $x=19$ which is prime.

the only pair satisfying the conditions is (19,3)

$x+y= 19+3 = \boxed{22}$