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.
This is an interesting question that I've wondered about myself, so I can't really answer it properly but I'll make a couple elementary observations. First, for a lattice in ${\mathbb Z}[\tau]\subset{\mathbb C}$ to be an ideal just means that multiplication by $\tau$ takes the lattice to itself. For example, for the Gaussian integers this says the lattice is invariant under 90 degree rotation about the origin. It is easy to see that this means the lattice is a square lattice with a basis consisting of two of its shortest vectors. (This is really just a disguised version of the Euclidean algorithm for Gaussian integers.) Thus every ideal is principal in this case.
Back to the general case, another observation is that if two ideals are in the same ideal class, then the two lattices are related by an orientation-preserving similarity, that is, rotation and rescaling, which is what multiplying an ideal by an element of ${\mathbb Z}[\tau]$ does to a lattice. (It seems the converse should be true as well). For example in the case $\tau =\sqrt5i$ the principal ideal class consists of rectangular lattices similar to ${\mathbb Z}[\tau]$ itself, and the other ideal class (the class number is 2 here) consists of lattices similar to the lattice $(2,1+\sqrt5i)$ which is skewed rather than rectangular. It is enlightening to draw a picture to see how this lattice is invariant under multiplication by $\sqrt5i$.
Another thing that can be viewed geometrically is the correspondence between ideal classes and equivalence classes of binary quadratic forms of fixed discriminant. An ideal, viewed as a lattice, determines a quadratic form by restricting the usual norm (squared) $x^2 + y^2$ to the lattice, then renormalizing suitably to make equivalent ideals have equivalent quadratic forms. A textbook that explains this, to some extent at least, is Advanced Number Theory by Harvey Cohn.
It would be interesting to work out some more examples to see what the different similarity classes of lattice-ideals look like, especially in cases when the class number is larger than 2. Is the structure of the ideal class group somehow visible in how the similarity classes are related?
Best Answer
This is false. The smallest counterexample is $d = 34$. Let $K = \mathbb{Q}(\sqrt{34})$. The fundamental unit in $\mathcal{O}_{K} = \mathbb{Z}[\sqrt{34}]$ is $35 + 6 \sqrt{34}$, which has norm $1$, and therefore, there is no element in $\mathcal{O}_{K}$ with norm $-1$.
However, $\frac{3}{5} + \frac{1}{5} \sqrt{34}$ has norm $-1$, so there is an element of norm $-1$ in $K$.