TITLE: Elliptic curve with same number of points over two different fields
QUESTION [6 upvotes]: Following a discussion with a number theory professor, we arrived at the following question : 
Can we find an elliptic curve (short form : $y^2=x^3+ax+b$) with an identical number of points on two different finite fields: $GF(q)$ and $GF(q^r)$ with $q$ prime.
Looking at some brute force results, we found only one result : 
the elliptic curve defined by $y^2=x^3+2x+1$ with fields of order 3 and 9.
We only looked at first 100's primes $q$, with power up to 10 so far, but we cannot understand what's specific about this choice, and why is this the only solution we can find so far...
any idea? Any clue as how we could attack this problem?
Thanks
Edit : Using Weil conjecture, we can start with, $\forall q$, $\exists \alpha \in \mathbb{C}$ such that $\forall r$ : 
$$ \#E(\mathbb{F}_{q^r})= (q^r+1) - (\alpha^r+\bar{\alpha}^r) $$
with $\alpha$ and $\bar{\alpha}$ conjugate, such that $\mid\alpha\mid=\sqrt{q}$

REPLY [2 votes]: There are no examples for $q>4$, of elliptic curves $E$ over $\mathbb{ F}_q$ such that $E(\mathbb{ F}_q)=E(\mathbb{ F}_{q^r})$ for some $r\ge 2$. 
Proof
Using the Hasse-Weil bounds, that can be deduced from Hasse's theorem that you stated, we have
$$ |E(\mathbb{ F}_q)| \le q+1 + 2 \sqrt{q}$$
and
$$ |E(\mathbb{ F}_{q^r})| \ge q^r+1 - 2 \sqrt{q^r}\ge q^2+1-2q$$
if $r\ge 2$ (as $q>1$). Hence
 $$ |E(\mathbb{ F}_{q^r})|- |E(\mathbb{ F}_q)| \ge q^2-3q- 2\sqrt{q}=\sqrt{q}(\sqrt{q}+1)^2(\sqrt{q}-2)
$$
which is $ >0$ if $q>4$.