I answer the original question, about the existence of infinitely many $D$'s, $D\equiv 5 \pmod{8}$, such that the quadratic $D$-twist of $E: y^2=x^3+17x$ has infinitely many points. As indicated, this should imply under the parity conjecture, that there are infinitely many such twists with rank $\ge 2$ (and even).

First of all, about the torsion of the twists of the curve $E$: it is always the group with two elements (over the rationals). The torsion point is $(0,0)$. This can be deduced by showing that the twists cannot contain a point of order 4 or a point of order 6.

So we only need to show there are infinitely many $D$'s with a rational point with $x\ne 0$ in $E_D$ and $D\equiv 5 \pmod{8}$.

Note that one can do a quadratic twist with any non-zero rational number $z$, but there is a unique $D$ square free integer such that $z/D$ is a square; then the quadratic twist with respect to $z$ and with respect to $D$ are isomorphic.

Consider the polynomial $f=y^2-x^3-17z^2x$ where the variables are $x$, $y$ and $z$, over the field of rational numbers. This defines an affine surface which is rational and easily parametrizable (the points with $x=0=y$ are special, but we are not interested with them).

A possible parametrization is $x=t^2/(s^2 + 17)$, $y=t^3/(s^3 + 17s)$ and $z=t^2/(s^3 + 17s)$. Now your question is equivalent to the question: there exists infinitely many integers $D$, congruent to $5$ modulo 8, such that
$D= u^2t^2/(s^3 + 17s)$ for some $u$, $t$ and $s\in \mathbb{Q}$? It is clear that this is equivalent to just ask if
$D$ can be written as $D= u^2(s^3 + 17s)$ for some $u$, $s\in \mathbb{Q}$?

Or, more directly and elementarily, one can see that the point $(x,y)=(1/(s^2 + 17),1/(s^3 + 17s))$ is a solution of the equation
$y^2=x^3+17/(s^3 + 17s)^2x$.

Writting $s=\frac ab$, with $a$ and $b$ integers, then $s^3 + 17s$ is equal to $b(a^3+17ab^2)$ modulo squares. Notice that if $b(a^3+17ab^2)$ is congruent to $5$, and $D$ is a square free integer equal to $b(a^3+17ab^2)$ modulo squares, then $D\equiv 5 \pmod{8}$. But this never happens, as $b(a^3+17ab^2)$ is always even.

Hence, we can look for $b(a^3+17ab^2) \equiv 20 \pmod{32}$, which will imply that, if $D$ is a square free integer equal to $b(a^3+17ab^2)$ modulo squares, then $D\equiv 5 \pmod{8}$.

But it is easy to find infinitely many solutions of this equation: for example, if $b\equiv 4\pmod{32}$ and $a\equiv 5 \pmod{32}$.

p.s. It could be possible to prove that there are infinitely many quadratic twists of this curve that have at least two independent points, and may be deduce the same for the quadratic twists with the condition you are interested (without using any conjecture), using the ideas given by Mestre and by Steward and Top. See the paper "On Ranks of Twists of Elliptic Curves and Power-Free Values of Binary Forms" and the references therein.

infinitely many twists $E_D$, the parameter you can vary is $D$, right? But in your post, $D$ seems to be fixed... $\endgroup$5more comments