Elliptic curve with same number of points over two different fields
In characteristics two and three you don't have the short Weierstrass form, and this leaves room for more examples. Below there are examples of
- $q=2$, $r=2$,
- $q=2$, $r=3$,
- $q=4$, $r=2$.
Consider the elliptic curve $$ y^2+y=x^3+x $$ defined over $\Bbb{F}_2$. It is relatively easy to see that all the solutions with $x,y\in\Bbb{F}_4$ actually have $x,y\in\Bbb{F}_2$. Basically because $y^2+y=0$ when $y\in \Bbb{F}_2$, and $y^2+y=1$ when $y\in\Bbb{F}_4\setminus \Bbb{F}_2$, but $x^3+x\neq0,1$ when $x\in\Bbb{F}_4\setminus\Bbb{F}_2$.
An explanation in terms of Hasse-Weil formula is to observe that $\# E(\Bbb{F}_2)=5$, Every point $(x,y)\in\Bbb{F}_2\times\Bbb{F}_2$ is a solution, and then we have the point at infinity. Therefore $$ \alpha+\overline{\alpha}=\alpha+\frac2\alpha=-2. $$ Squaring this equation gives $$ \alpha^2+\overline{\alpha^2}=\alpha^2+\frac4{\alpha^2}=(\alpha+\frac2\alpha)^2-4=(-2)^2-4=0. $$ Therefore $\# E(\Bbb{F}_4)=4+1-0=5=\# E(\Bbb{F}_2).$
Other characteristic two examples are the curves $$ y^2+xy=x^3+1 $$ and $$ y^2+xy=x^3+x. $$ Both have four points over $\Bbb{F}_2$ but no other solutions over $\Bbb{F}_8$.
From $\#E(\Bbb{F}_2)=4$ we can solve $\alpha=(-1+i\sqrt7)/2$. Then $\alpha^3=(5-i\sqrt7)/2$ and thus, by Hasse-Weil $$ \# E(\Bbb{F}_8)=8+1-(\alpha^3+\overline{\alpha^3})=8+1-5=4. $$ An alternative explanation comes from the fact that the trace of $(x+\dfrac1x)$ is equal to $1$ for all $x\in\Bbb{F}_8\setminus\Bbb{F}_2$. The solvability criterion for having a solution $y\in\Bbb{F}_8$ would dictate this trace to vanish. Hence, no new points.
Yet another characteristic two example.
Consider the curve $$y^2+y=x^3.$$ With $x$ ranging over the field $\Bbb{F}_4$ we have $x^3\in\Bbb{F}_2$, therefore two solutions for $y$ to each $x$ and therefore nine points altogether. In this maximal case we must have $\alpha=\overline{\alpha}=-2$. But, then $\alpha^2=\overline{\alpha}^2=4$, so over the field $\Bbb{F}_{16}$ we have $16+1-(\alpha^2+\overline{\alpha}^2)=9$ points also. The trace condition for solvability of a quadratic leads to the same conclusion. After all, the cube of an element $x\in\Bbb{F}_{16}\setminus\Bbb{F}_4$ is of order five, and those all have trace $1$.
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 $q>4$.