There is a hint to show that if there were points in $\mathbb{Q}$, then there would exist $r,s, t \in \mathbb{Z}$ with $\gcd(r, t) = 1$ such that $2s^2 = t^4 - 17r^4$ , and then show that any prime dividing $s$ is a quadratic residue modulo $17$.
I'm stuck on the first bit - if you suppose $\left(\frac{a}{b}, \frac{c}{d}\right)$ is a point on the curve in $\mathbb{Q}$ (so $a, b, c, d \in \mathbb{Z}$) then you get to $2a^2d^4 = b^2c^4 - 17b^2d^4$, but I'm not sure where to go from here since you don't know that $b$ is a square?
$b$ need not be a square. Instead, the trick is to show that you can actually divide the equation by $b^2$.
Let $p$ be a prime that divides $b$ and is not 2. Let $p^n$ be the highest power of $p$ that divides $b$.
Since $\gcd(a,b) = 1$, and we know that $ 2a^2 d^4 \equiv 0 \pmod{p^{2n}}$, thus $ d^4 \equiv 0 \pmod{p^{2n}}$ and this implies that $ p^n \mid d^2 $.
If $ 2^n \mid \mid b$ (highest power of 2 that divides $b$), then we have $ 2 \not \mid a$. Similarly, we get $ 2 d^4 \equiv 0 \pmod{2^{2n}}$. Let $2^m \mid \mid d^2$, then we know that $ 1 + 2m \geq 2n$. Since $m$ and $n$ are integers, we can conclude that $m \geq n$. Thus, we have $ 2^n \mid d^2$.
Hence, $b \mid d^2$, and so we may divide throughout by $b^2$, to obtain $ 2 \left( a \times\frac{d^2}{b} \right)^2 = c^4 - 17 d^4$, and $\gcd (c, d) = 1 $ by assumption.
In general, you see that this will work, as long as the coefficient (2) is square-free.