Mostly, you should look at a number of items at
(address updated in 2018):
http://zakuski.math.utsa.edu/~kap/
including Dickson_Diagonal_1939.pdf and Kap_Jagy_Schiemann_1997.pdf to begin with.
Now that I think of it, you also need to read Kap_All_Odd_1995.pdf at the same place, also a new preprint by Jeremy Rouse on his 451 Theorem(s), as it is readily decided whether a form represents the single number 2. In a different direction, you need Wai Kiu Chan and Byeong-Kweon Oh, "Positive Ternary Quadratic Forms with Finitely Many Exceptions," Proceedings of the A.M.S., Volume 132, Number 6, Pages 1567-1573 (2004).
The overriding fact is that ternary forms, like binary forms, are collected together in genera. Unlike binary forms, these vary in size (class number) for a fixed discriminant. The good thing is the result of Jones, every number given by congruence conditions (a finite number of "progressions") is, in fact, represented by at least one form in the genus.
So, an example, $x^2 + 4 y^2 + 9 z^2$ is not regular, so it is not in Dickson's list. The genus of this form represents all numbers not of shape $9 n \pm 3, \; 8 n + 3, \; 4^k (8n+7).$ The other class in the genus is $x^2 + y^2 + 36 z^2.$ Between the two forms, all eligible numbers are represented. It is not difficult to prove that $x^2 + 4 y^2 + 9 z^2$ misses only the single number 2 out of the eligible numbers. So, if you were so minded, you could say that $x^2 + 4 y^2 + 9 z^2$ represents all primes that pass the above restrictions as well as not being $0 \pmod 2.$ As the restriction $9n \pm 3$ does not affect larger primes, just larger composite numbers, one could also say that $x^2 + 4 y^2 + 9 z^2$ represents all primes $p \equiv 1 \pmod 4.$
So there is a built in problem with your formulation. If one of your positive ternaries $x^2 + m y^2 + n z^2$ has finitely many exceptions, in particular finitely many prime exceptions $p_1,p_2,\ldots,p_k,$ the form can be said to represent every prime $p$ fitting the original restrictions and the new restrictions $p \neq 0 \pmod {p_k.}$ As a result, your list of ternary forms is infinite and unprovable. As Franz says, you need a tighter formulation.
If you like, email me your list of forms, we can discuss it.
A better example of the possible horror: Ono and Soundararajan (1997) showed that Ramanujan's form $x^2 + y^2 + 10 z^2$ has only squarefree numbers as sporadics (numbers represented by some form in the genus but not by this form, "exceptions" for Chan and Oh). They also showed that GRH implies that the known list is complete. So, GRH implies that $x^2 + y^2 + 10 z^2$ represents all primes not divisible by any of 3, 7, 31, 43, 67, 79, 223, 307, 2719. The other sporadics are composite. At the same time, it is easy to show that the form represents all numbers $n \equiv 5 \pmod 6,$ first pointed out in a letter from J.S.Hsia to Kaplansky, later a cheap proof by me, and one by Oh.
EDIT: it occurs to me that an alternate property could be: a positive ternary form will be defined to be fungible if its sporadics are all composite. Or perhaps funicular. I looked it up, the best would be frangible.
EDIT TOOOOO: I thought I might find examples of forms $x^2 + m y^2 + n z^2$ that seem to be fungible, or perhaps funicular, or frangible, despite lacking proof. The first example is $$ x^2 + y^2 + 48 z^2 \neq 21 \cdot 9^k $$ compared with the other form in that genus, $2 x^2 + 2 y^2 + 13 z^2 + 2 y z + 2 z x,$ checked on numbers up to 1,250,000. Very similar, $$ x^2 + 4y^2 + 20 z^2 \neq 77 $$ compared with the other form in that genus, $4 x^2 + 4y^2 + 5 z^2,$ also checked on numbers up to 1,250,000. In this second case it is easy to show that each form of the genus represents 4 times any number represented by the other form, and no numbers $2 \pmod 4$ are represented anyway, so only odd numbers come up. Anyway, 21 and 77 are composite. I have not proved these completely, just checked on computer.
EDIT TOOTOOTOO: I got an opinion from Jeremy Rouse. He points out that any positive ternary has two possible causes for having infinitely many numbers missed (compared to its genus), those being high divisibility by anisotropic primes or spinor exceptional classes. These two phenomena affect only finitely many squareclasses. In both cases, we do not increase the set of primes missed, with the result that a positive ternary fails to represent only a finite number of eligible (by congruence conditions) primes. This also explains, to some degree, the reference to Duke and Schulze-Pillot (1990). The final corollary says that any sufficiently large number that is primitively represented by some form in the same spinor genus is represented by the form of interest. There are only a few spinor exceptional squareclasses, so, even in an irregular spinor genus, we can only miss finitely many squarefree numbers, as those other than the spinor exceptional integers are represented by something in the same spinor genus, and primes are squarefree and therefore represented primitively if at all. I think I've caught up now. Note the D_S-P results give no effective bound, so we cannot identify the primes missed without some fortunate accident such as regularity, spinor regularity, regularity with regard to all odd numbers, and so forth.
Best Answer
You want the idoneal numbers, http://oeis.org/A000926 and http://en.wikipedia.org/wiki/Idoneal_number
See also pages 81-82 in Duncan A. Buell, Binary Quadratic Forms
Depending what you mean by 32, the primes represented by $x^2 + 8 y^2$ are, in fact, given by congruences. However, half of those same primes are represented by $x^2 + 32 y^2,$ while the other half are represented by $4 x^2 + 4 x y + 9 y^2.$ The condition saying which are which is not simply congruences.
EDIT: it is possible 32 can be finished by biquadratic reciprocity, in which case it is in print somewhere. For instance, given a prime $p \equiv 1 \pmod 4,$ there is a representation $p = x^2 + 64 y^2$ in integers if and only if $2$ is a fourth power modulo $p.$ In comparison, we get $p = x^2 + 14 y^2$ for $p \neq 2,7 $ if and only if $ ( -14 | p ) = 1$ and $ (x^2 + 1)^2 \equiv 8 \pmod p$ has an integer solution (Cox page 115).
Either way, there is a monic irreducible polynomial $f_{32}(z)$ of degree 4 (as $h(-128) = 4$) such that, if an odd prime $p$ does not divide the discriminant of $f_{32}(z),$ then we can write $p = x^2 + 32 y^2$ if and only if $(-2 | p) = 1$ and $f_{32}(z) \equiv 0 \pmod p$ has an integer solution. This is Cox, page 180, Theorem 9.2.
EDIT TOOO: What you want is Lemma 3.10 on page 333 of LIU_WILLIAMS Tamkang Journal of Mathematics, Volume 25, Number 4, Winter 1994. I am going to need to check in various ways, but I already think that $p$ is represented by $x^2 + 32 y^2$ if and only if $p \equiv 1 \pmod 8$ and $(z^2 - 1)^2 + 1 \equiv 0 \pmod p$ has a solution with an integer $z.$ Checking... Yes, this is correct. Theorem 4.1 on the same page, Table right below it. There is a bit of work showing that one root and $p \equiv 1 \pmod 8$ actually shows four linear factors.