Is $\mathbb{R}$ isomorphic to $\mathbb{R}(x)$?

Another argument.

In the field $\Bbb{R}$ every element $z$ or its additive inverse is a square of some element of the field. This property is preserved under an isomorphism. But the element $x\in\Bbb{R}(x)$ is not a square of any element. Neither is $-x$. Therefore we can conclude that $\Bbb{R}$ and $\Bbb{R}(x)$ are not isomorphic.

No. By adjoining to $\Bbb{R}$ a root of the equation $T^2=-1$ we get an algebraically closed field. If $\Bbb{R}$ and $\Bbb{R}(x)$ were isomorphic, the same would happen by adjoining a root of $T^2=-1$ to $\Bbb{R}(x)$ (observe that any isomorphism maps $-1$ to itself). But $\Bbb{R}(x)[i]\simeq\Bbb{C}(x)$ is not algebraically closed, so this is not the case.

$\mathbb R$ satisfies the condition $\forall t\exists s(t=s^2\lor -t=s^2)$, but $\mathbb R[x]$ does not (consider $t=x$). So they're not isomorphic (in fact they're not even "elementarily equivalent").