There is a slightly different proof which works over any field $k$ (of characteristic different from 2 and 3). The first field is just $k(x,y)$. As Rita indicated, if it is isomorphic to the second field as $k$-extension, than there exists a birational map $f: \mathbb P^2\to E$. This birational map induces a birational map over the algebraic closure of $k$, so we can suppose $k$ is algebraically closed. By two arbitrary points $p, q$ where $f$ is defined, it passes a line $L$. As $E$ is not rational, by Lüroth $f|_L$ is constant. So $f$ is constant, contradiction. More quickly, one can say that the existence of $f$ implies that $E$ is unirational and this is impossible.

There is a slightly different proof which works over any field $k$ (of characteristic different from 2 and 3). The first field is just $k(x,y)$. As Rita indicated, if it is isomorphic to the second field as $k$-extension, than there exists a birational map $f: \mathbb P^2\to \mathbb P^1\times E$. This birational map induces a birational map over the algebraic closure of $k$, so we can suppose $k$ is algebraically closed. Composing with the projection to $E$, we get a dominant rational map $g: \mathbb P^2\to E$. By two arbitrary points where $g$ is defined, it passes a line $L$. As $E$ is not rational, by Lüroth $g|_L$ is constant. So $g$ is constant, contradiction. More quickly, one can say that the existence of $g$ implies that $E$ is unirational and this is impossible.

EDIT: the rational map $\mathbb P^2\to E$ is not birational ! but dominant.

There is a slightly different proof which works over any field $k$ (of characteristic different from 2 and 3). The first field is just $k(x,y)$. As Rita indicated, if it is isomorphic to the second field as $k$-extension, than there exists a birational map $f: \mathbb P^2\to E$. This birational map induces a birational map over the algebraic closure of $k$, so we can suppose $k$ is algebraically closed. By two arbitrary points $p, q$ where $f$ is defined, it passes a line $L$. As $E$ is not rational, by Lurôth $f|_L$ is constant. So $f$ is constant, contradiction. More quickly, one can say that the existence of $f$ implies that $E$ is unirational and this is impossible.

There is a slightly different proof which works over any field $k$ (of characteristic different from 2 and 3). The first field is just $k(x,y)$. As Rita indicated, if it is isomorphic to the second field as $k$-extension, than there exists a birational map $f: \mathbb P^2\to E$. This birational map induces a birational map over the algebraic closure of $k$, so we can suppose $k$ is algebraically closed. By two arbitrary points $p, q$ where $f$ is defined, it passes a line $L$. As $E$ is not rational, by Lüroth $f|_L$ is constant. So $f$ is constant, contradiction. More quickly, one can say that the existence of $f$ implies that $E$ is unirational and this is impossible.