These are exactly the primes where the Neron model of $E$ has fiber type, in Kodaira's table, anything other than $I_0$ and $I_0^*$. Here $I_0$ represents good reduction and $I_0^*$ represents a quadratic twist of good reduction by a ramified extension (quadratic twisting by an unramified extension preserves the reduction type).
(At least, this is true away from primes above 2, which are surely more complicated).
The fiber type can be computed by Tate's algorithm, but to tell whether the type is $I_0$ or $I_0^*$, you can simply perform a ramified quadratic twist and check whether either the curve or its quadratic twist has good reduction. The number you quadratic twist by doesn't matter, as long as it is $\pi$-adic valuation, where $\pi$ is the prime of bad reduction, is odd.
For $\pi$ in $\mathbb Q(\sqrt{-5})$ lying over a prime $p$ of $\mathbb Q$, unless $\pi$ is ramified (i.e. $\pi$ lies above $5$), then any number which has $p$-adic valuation odd will do the trick.
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 $D \equiv 5 \pmod 8$. $\square$
Best Answer
You can compute the root number over $\mathbb{Q}$ of any elliptic curve of the form $y^2=x^3+\alpha x$ for $\alpha\in \mathbb{Q}^*$ using the formulae in section 4B in Density of rational points on isotrivial rational elliptic surfaces, by Anthony Várilly-Alvarado.
In your case $\alpha=17D^2$, for $D$ odd and square free (if $p^2$ divides $D$, then the elliptic curve $E_D$ and $E_{D/p^2}$ are isomorphic), the root number is $1$ if and only if $D\equiv 1 \pmod 4$ (it is not true what you said about any odd integer $D$). In fact we will show that $$ W(E_D)=\left(\frac{-1}{D} \right) $$
Using the formulae there, we need to compute first what he calls $W_2(17D^2)$ (and it is the root number at $2$). It is always $-1$, as $v_2(17D^2)=0$ and $17D^2 \equiv 1, 9,15 \pmod{16}$.
Then we compute $W_3(17D^2)=-1=\left(\frac{-1}{3} \right)$ if and only if $3$ divides $D$ (but 9 doesn't, as $D$ is square-free).
Now, following his notation, the number $$R(17D^2)=\left(\frac{-1}{17D^2} \right)W_2(17D^2)W_3(17D^2)(-1)^{v_3(D^2)}=-\left(\frac{-1}{17} \right)W_3(17D^2)=-W_3(17D^2).$$
Now, for the contributions of the other primes, we need to consider only the primes $p>3$ that divide $D$, and then $p^2$ divides $\alpha=17D^2$, and $v_p(\alpha)=v_p(17D^2)=2$, as $p^2$ doesn't divide $D$.
So, we get finally that the root number is $$W(17D^2)=-R(17D^2)\prod_{p\mid D,p>5} \left(\frac{-1}{p} \right) =W_3(17D^2)\prod_{p\mid D,p>5} \left(\frac{-1}{p} \right)=\prod_{p\mid D} \left(\frac{-1}{p} \right) =\left(\frac{-1}{D} \right) ,$$ which is $1$ if and only if $D\equiv 1 \pmod 4$.
Another more direct proof, but valid only if $17$ does not divide $D$, is by using the fomulae in Root number of a quadratic twist of an elliptic curve, which in fact is explained in of Rubin's and Silverberg's article Twists of elliptic curves of rank at least four. As the conductor $N$ of $E$ given by $y^2=x^3+17x$ is $N=2^617^2$, the root number is $$W(E_D)=\left(\frac{-N}{D} \right)W(E)=\left(\frac{-1}{D} \right)W(E)=\left(\frac{-1}{D} \right),$$ as the root number of $E$ is 1, if $D$ is odd and not divisible by $17$.