The answer is yes. As was observed, the condition of being pre-Pell is simply the stipulation that $k$ is a sum of two squares: that is, if $p | k$ and $p \equiv 3 \pmod{4}$ then $p$ must divide $k$ with even multiplicity. If we assume $k$ is square-free, then it is divisible only by $2$ or primes of congruent to $1 \pmod{4}$.

In fact, if $k$ is pre-Pell then *all* of its odd prime divisors must be congruent to $1 \pmod{4}$. To see this, suppose $x^2 + 1$ is divisible by $p^2$ for some $p \equiv 3 \pmod{4}$. Then in particular it must be divisible by $p$, so the congruence $x^2 + 1 \equiv 0 \pmod{p}$ must be soluble. But this is not the case: $-1$ is a square mod $p$ if and only if $p = 2$ or $p \equiv 1 \pmod{4}$.

We now work on the equivalences. Suppose that $k$ is such that there exists $k^\prime$ such that the negative Pell equation is soluble for both $k^\prime$ and $kk^\prime$. Being a sum of two squares is a necessary condition for negative Pell to be soluble, hence the latter condition implies that $k = (kk^\prime)/k^\prime$ must also be a sum of two squares. To see this, suppose that $kk^\prime$ has a prime divisor $p$ congruent to $3 \pmod{4}$ (otherwise it is obvious that $k$ is a sum of two squares). Suppose that $p^{2m} || kk^\prime$ and $p^{2n} || k^\prime$. Then $p^{2m-2n} || k$, hence $p$ divides $k$ to even multiplicity. Therefore, $k$ is pre-Pell as desired.

The converse is much harder. Suppose that $k$ is a sum of two squares. We can use the fact that for any prime $p \equiv 1 \pmod{4}$, the negative Pell equation $x^2 - py^2 = -1$ is soluble. Thus we may reduce the question to the following: does there exist a prime $p \equiv 1 \pmod{4}$ such that $x^2 - kpy^2 = -1$ is soluble?

Fortunately this can be done via *governing fields*: that is, for each $k$ there exists a number field $F_k$ such that the 4-rank of the class group of the field $\mathbb{Q}(\sqrt{kp})$ is determined by the splitting behaviour of $p$ in $F_k$. In particular, if the 4-rank of the class group of $\mathbb{Q}(\sqrt{kp})$ is zero, then negative Pell is soluble. We are then done by Chebotarev's density theorem guaranteeing the existence of such a prime (in fact, infinitely many).