Can you show rank E(Q) = 1 exactly for infinitely many elliptic curves E over Q without using BSD? Let $K$ be a number field and let $\mathcal O_K$ be the ring of integers. Following this paper of Cornelissen, Pheidas, and Zahidi, a key ingredient needed to show that Hilbert's tenth problem has a negative solution over $\mathcal O_K$ is an elliptic curve $E$ defined over $K$ with rank$(E(K))=1$.
Recently Mazur and Rubin have shown that such a curve exists assuming the Shafarevich-Tate conjecture for elliptic curves over number fields. They actually use a weaker, but still inaccessible hypothesis (See conjecture $ШT_2$).
If you wanted to eliminate the need for this hypothesis you would have to write a proof that simultaneously demonstrated that rank$(E(K))=1$ for infinitely many pairs $(K,E)$ where $E$ is an elliptic curve defined over $K.$ This raises (as opposed to begs) the easier question:

Can you show unconditionally that rank$(E(\Bbb Q)) = 1$ for infinitely many elliptic curves $E$ over $\Bbb Q$?

It would appear that Byeon, Jeon, and Kim have done so in this paper (probably need an institutional login). Vatsal obtains a weaker result here that still does the job. Unfortunately both of these results invoke the fact that the BSD rank conjecture is true for elliptic curves over $\Bbb Q$ with analytic rank 1. Which won't help at present working over number fields.

Can anyone do the above WITHOUT invoking the proven part of the BSD rank conjecture or assuming any conjectures?

 A: This is just a tiny follow up to Victor Miller's remark above.  Manjul Bharghava has a second paper with Shankar entitled "Ternary cubic forms having bounded invariants, and the existence of a positive proportion of elliptic curves having rank 0", and in there he states the following theorem:
Theorem 5: Assume $Sha(E)$ is finite for all $E$. When all elliptic curves $E/\mathbf{Q}$ are ordered by height, a positive proportion of them have rank $1$. 
This obviously doesn't answer your original question, because Manjul is assuming that Sha is finite.  However, I believe his arguments for this theorem use results toward BSD over $\mathbf{Q}$. 
EDIT: Crud -- I meant to say that I believe his arguments don't use results toward BSD, which was the whole point.  Oops. 
A: If $K$ is totally real, then the rank part of BSD is known in analytic rank 1 (using CM points for instance) for modular elliptic curves. As potential modularity is also known under the same hypotheses, I am quite confident that this part of Vatsal and Byeon, Jeaon and Kim generalizes. Is it OK for you to assume that $K$ is totally real?
On the other hand, their proofs also use some results on the class group of quadratic fields, so this part would not generalize (as far as I know, which is not very far).
A: I don't think that this should be too hard: take a simple family of curves, such as 
$y^2 = x^3 + px$ or something similar, and choose $p$ from a certain set of residue classes to guarantee that the 2-Selmer group has rank 1. You can complete the proof either by invoking rather deep constructions using Heegner points, or finding a family for which conditions such as $p = a^4 + b^2$ (there are infinitely many primes of this form) give you a global point (choose your family in such a way that $b^2 = px^4 - y^4$ occurs as a 
principal homogeneous space in your standard 2-descent; see e.g. Silverman's book).
Sorry for being a little bit vague - a hard disk crash currently prevents me from looking at my own notes.
