Let me turn my comment into an answer. There are indeed infinitely many such twists with a non-torsion point. By Nagell–Lutz, it suffices to produce infinitely many different squarefree integers $D \equiv 5 \pmod 8$ for which there exist $x,y \in \mathbf Q$ such that $y^2 = x^3+17D^2x$ and $x$ and $y$ are not both integers. Writing $X = Dx$ and $Y = D^2y$, we get the equivalent equation $DY^2 = X^3+17X$ (but beware that the Nagell–Lutz criterion does not apply in these coordinates).
One way to produce values of $D$ is as follows: there are infinitely many primes $p > 5$ such that $-17$ is a square modulo $p$. For such a prime $p$, pick $m \in \mathbf N$ such that $v_p(m^2+17) = 1$ and $m \equiv 5 \pmod 8$. Pick $n \in \mathbf N$ with $4n \equiv 1 \pmod p$ and $n \equiv 1 \pmod 8$. Set $X = \tfrac{m}{4n}$, and write $X^3+17X$ as $DY^2$ with $D \in \mathbf Z$ squarefree and $Y \in \mathbf Q$. Finally, set $x = \tfrac{X}{D}$ and $y = \tfrac{Y}{D^2}$.
We claim that $(x,y)$ is a non-torsion point on $E_D$, that $p \mid D$, and that $D \equiv 5 \pmod 8$. Since we can carry out this process for infinitely many primes $p$, we conclude that the numbers $D$ obtained this way also form an infinite set. It is clear that $x$ is not an integer, so $(x,y)$ cannot be a torsion point.
To see that $p \mid D$, note that $v_p(DY^2) = v_p(X^3+17X) = v_p(X^2+17) = 1$, so $v_p(D) = 1$ by definition of $D$. To see that $D \equiv 5 \pmod 8$, note that
$$DY^2 = X^3+17X = X(X^2+17) = \tfrac{m}{4n}\cdot \tfrac{m^2+272n^2}{16n^2}.$$
Since $m$ and $n$ are odd, we see that $v_2(Y^2) = -6$ and $v_2(D) = 0$. Factoring out all even denominators, the above equation reads
$$D \cdot (2^3Y)^2 = \tfrac{m}{n} \cdot \tfrac{m^2+272n^2}{n^2}.$$
Since $2^3Y$ is a unit in $\mathbf Z_{(2)}$, its square is 1 modulo 8. Since $m$ and $n$ are odd, we get $\tfrac{m^2+272n^2}{n^2} \equiv 1 \pmod 8$, and we have $\tfrac{m}{n} \equiv 5 \pmod 8$ by the choice of $m$ and $n$. Putting everything together gives $D \equiv 5 \pmod 8$. $\square$
There is a formula but it involves both cubic twists. Let $E: y^2 = x^3+B$ be an elliptic curve over $\mathbb{Q}$ with $j=0$ as the one in the question. Let $D$ be a cubefree integer. Set $E_1: y^2=x^3+D^2\,B$ and $E_2:y^2=x^3+D^4\,B$.
Both $E_1$ and $E_2$ become isomorphic to $E$ over $L=\mathbb{Q}\bigl(\sqrt[3]{D}\bigr)$. I believe the formula is
$\DeclareMathOperator{\rk}{rk}$
$\rk E(L) = \rk E(\mathbb{Q}) + \rk E_1(\mathbb{Q}) + \rk E_2(\mathbb{Q}).$
This can easily checked to be the case for the given curve and some small $D$.
Let $K=L(\mu_3)$ be the Galois closure of $L$ and let $F=\mathbb{Q}(\mu_3)$. Denote by $G$ the cyclic Galois group of $K/F$ and write $\chi$ and $\bar \chi$ for the two non-trivial characters of $G$.
First the motivation: The $L$-function associated to the $G_F$-representation $T_p E\otimes \mathbb{C}[G]$ is the $L$-function of $E/K$. Let $\psi$ be the Grössencharacter associated to $E$, then this $L$-function splits into six $L$-functions, namely those for $\psi$, $\bar\psi$, $\psi\chi$, $\overline{\psi\chi}$, $\psi\bar{\chi}$, and $\bar{\psi}\chi$. The first two give the $L$-function of $E/F$, the middle two the $L$-function of $E_1/F$ and the last two the $L$-function of $E_2/F$. The Birch and Swinnerton-Dyer conjecture now implies that $\rk E(K) = \rk E(F)+\rk E_1(F)+\rk E_2(F)$.
We can prove this directly by looking at the $G$-action on $E(K)$. Write $g$ for the element of $G$ that sends $\alpha\in L$ with $\alpha^3 =D$ to $\zeta\cdot \alpha$ where $\zeta^3=1$. Let $[\zeta]\in \operatorname{End}(E)$ be the element of order $3$ given by $[\zeta](x,y) = (\zeta \cdot x, y)$. Define $E(K)_i = \bigl\{ P \in E(K)\, :\, g(P) = [\zeta]^i(P)\bigr\}$. Then $E(K)_0 = E(F)$ and the map $E_1(F) \to E(K)$ sending $(x,y)$ to $(x/\alpha^2, y/\alpha^3)$ has image equal to $E(K)_1$. Similar $(x,y)\mapsto (x/\alpha^4,y/\alpha^6)$ brings $E_2(F)$ to $E(K)_2$.
The kernel of the map from $E(F)\oplus E_1(F)\oplus E_2(F)$ to $E(K)$ is contained in the finite subgroup $E(F)[\zeta-1]$. For any $P\in E(K)$ we have $3P = (1+g+g^2)(P) + (1+[\zeta]g+[\zeta]^2g^2)(P) + (1+[\zeta]^2g+[\zeta]^4g^2)(P)$ which belongs to the sum $E(F)+E(K)_1+E(K)_2$. Thherefore the cokernel is also finite. Hence $\rk E(K) = \rk E(F)+\rk E_1(F)+\rk E_2(F)$.
Since $E$ has complex multiplication by an order in $F$, the rank of $E(F)$ is twice the rank of $E(\mathbb{Q})$; and similarly for $E_1$ and $E_2$. The representation $E(K)\otimes\mathbb{C}$ of the Galois group (equal to $S_3$) of $K/\mathbb{Q}$ splits as $\mathbb{1}^a + \epsilon^a+\rho^b$ where $\epsilon$ is the irreducible non-trivial $1$-dimensional and $\rho$ is the irreducible $2$-dimensional representation of $S_3$. Therefore
$\rk E(L) = a+b = \tfrac{1}{2} (2a+2b) = \tfrac{1}{2} \rk E(K) = \tfrac{1}{2}\cdot \bigl( \rk E(F) + \rk E_1(F) + \rk E_2(F) \bigr) =\rk E(\mathbb{Q}) + \rk E_1(\mathbb{Q}) + \rk E_2(\mathbb{Q})$.
Best Answer
Perhaps a few remarks too long to fit in a comment will help clarify the issue.
The question of whether there exist elliptic curves of arbitrarily large rank has a surprisingly long and interesting history. Poincare was the first to ask the question as to whether the rank of elliptic curves is uniformly bounded (in his time, it was not even known if the group of rational points of an elliptic curve is always finitely generated; this was ultimately proved by Mordell in 1922, 10 years after Poincare's death). For a long time no one could construct curves of all but the smallest ranks, leading to the opinion that the rank should be bounded to prevail, but towards the middle of the 20th century curves of double digit ranks started to appear. The winds changed leading to people to believe that the rank should be unbounded. The capstone of this trend was the Shafarevich-Tate theorem asserting that the rank of elliptic curves over a function field is in fact unbounded. This trend would persist into the 21st century, until a large number of different heuristics, ultimately united by work of Bhargava-Kane-Lenstra-Poonen-Rains (BKLPR). This was convincing enough, including top experts like Andrew Wiles*, that in fact the rank perhaps should be bounded after all.
Since the question at hand concerns the quadratic twist family of a single elliptic curve, it is useful to compare the analogous story for average rank of elliptic curves. Here there was hardly ever a doubt that this quantity ought to be bounded. Brumer first proved, in 1992, that assuming both the Birch/Swinnerton-Dyer conjecture and the Grand Riemann Hypothesis for elliptic curve $L$-functions, that the average rank of elliptic curves is in fact bounded. This was sharpened subsequently by Heath-Brown and Young who improved on the quantitative bounds. Earlier work of Silverman also confirmed boundedness for certain thin algebraic families. In 2010, groundbreaking work by Bhargava and Shankar removed all unproven hypotheses from the proof, and they showed unconditionally that when ordered by naive height, the average rank of elliptic curves is no more than $1.5$ (Binary quartic forms having bounded invariants, and the boundedness of the average rank of elliptic curves). This bound has subsequently been improved by obtaining averages for the 3-Selmer and 5-Selmer ranks of elliptic curves, respectively.
The Bhargava-Shankar theorem is the most significant progress towards the Katz-Sarnak conjecture, which asserts that the average rank of elliptic curves when ordered by any reasonable height (including by discriminant, conductor, or by naive height) ought to be $1/2$. Bhargava and Shankar showed that their conjecture follows from obtaining the exact average of $n_k$-Selmer ranks for any infinite sequence of $(n_k)$. The Katz-Sarnak conjecture is a refinement of an earlier conjecture of Goldfeld, which concerns quadratic twist families: it asserts that for any elliptic curve $E/\mathbb{Q}$, the average rank of the twist family $E_D : D \in \mathbb{N} \setminus \mathbb{N}^2$ ought to be $1/2$.
In 2016, groundbreaking work by Alex Smith almost confirmed Goldfeld's conjecture for curves with full rational 2-torsion. Subsequently, Smith has been able to generalize to other curves, including curves whose 2-torsion field is degree 6 over $\mathbb{Q}$, which is almost all curves. I believe what he proved is actually the statement that 100% of curves in the twist family have rank $0$ or $1$, but is unable to show that each is 50%.
Smith's techniques is very different from those used by Bhargava and Shankar, to the point of almost being incompatible. As such, it is not clear in the present setting that focusing on a single twist family is any easier than looking at all curves at once. Of course, both the Bhargava-Shankar and Smith theorems concern averages, which is far easier than uniform bounds.
*I attended a public lecture given by Sir Andrew Wiles in 2017, where he was given only 30 minutes to talk about his mathematical journey. Among other things he discussed his thoughts on the extended version of B-SD, and he spent the last five minutes talking about the BKLPR heuristics and how perhaps the rank might perhaps be bounded. That lecture was the turning point for me personally; prior to that I was of the opinion that the rank should be unbounded, and that the heuristics given, boiling down to the observation that the sum of the reciprocals of the conductors of curves with rank exceeding 22 ought to converge, does not preclude very thin families with rank going to infinity. However after that lecture I started entertaining the idea that the rank should be bounded, and after further evidence coming from relations to the uniform boundedness conjecture, that the rank is in fact bounded.