As rings, $\mathbb{Z}[x]$ is not isomorphic to $\mathbb{Q}[x]$ 
Theorem: $\mathbb{Z}[x]$ is not isomorphic to $\mathbb{Q}[x]$.

My attempt:

For a contradiction, suppose $\mathbb{Z}[x] \cong \mathbb{Q}[x]$. So there exists an isomorphism $f: \mathbb{Q}[x] \to \mathbb{Z}[x]$. Since $\mathbb{Q}$ and $\mathbb{Z}$ are integral domains, we can make the following assertions about the units of $\mathbb{Q}[x]$ and $\mathbb{Z}[x]$, denoted $\overline{\mathbb{Q}[x]}$ and $\overline{\mathbb{Z}[x]}$, respectively:
\begin{align*}
\overline{\mathbb{Q}[x]} = \overline{\mathbb{Q}} = \{q \in \mathbb{Q} \mid q \neq 0\} \\
\overline{\mathbb{Z}[x]} = \overline{\mathbb{Z}} = \{-1,1\}. 
\end{align*}
Since $f$ is a homomorphism, it maps units to units, so $f(\overline{\mathbb{Q}[x]}) \subset \overline{\mathbb{Z}[x]}$. Take $q_1, q_2, q_3$ distinct in $\overline{\mathbb{Q}[x]}$. Then $f(q_1), f(q_2), f(q_3)$ are units in $\mathbb{Z}[x]$, and since $f$ is injective, $f(q_1) \neq f(q_2) \neq f(q_3)$, meaning $\mathbb{Z}[x]$ has three distinct units, a contradiction.

How does this proof look?

Your proof looks good. Isomorphic rings have isomorphic groups of units.
Alternatively, the equation $2t=1$ has a solution in $\mathbb{Q}[x]$ but not in $\mathbb{Z}[x]$ and so the two rings cannot be isomorphic.
