= Solution
For $E:y^2=x^3+3$, direct counting gives
$$
\#E(\mathbb F_{17})=18,
\qquad
\#E(\mathbb F_{31})=43.
$$
Neither group order is divisible by $17$, so neither group contains a point of order $17$.
At $p=17$ the Frobenius trace is zero. The <elliptic-curve point count over a finite field> has trace recurrence
$$
a_0=2,\quad a_1=0,\quad a_n=-17a_{n-2},
\qquad
\#E(\mathbb F_{17^n})=17^n+1-a_n.
$$
For every $n\geq1$, this order is congruent to one modulo $17$. Consequently $E(\mathbb F_{17^n})$ has no point of order $17$ for any $n$.
At $p=31$, the trace is $a=31+1-43=-11$. On $E[17]$, Frobenius has characteristic polynomial
$$
X^2+11X+31\equiv X^2+11X+14\pmod{17}.
$$
Its discriminant is $14$, a nonsquare in $\mathbb F_{17}$, so its two distinct eigenvalues lie in $\mathbb F_{17^2}^{\times}$. Their orders divide $17^2-1=288$, whence $\pi^{288}=1$ on $E[17]$. Thus all of $E[17]$ is rational over $\mathbb F_{31^{288}}$, and in particular a point of order $17$ exists over some extension with $n\geq2$.
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