Identify the structure of $\mathbb{Z}[x]/(x^2+3,3)$ I assume that $\mathbb{Z}[x]/(x^2+3)\simeq\mathbb{Z}[i\sqrt{3}]$. Thus we should have $\mathbb{Z}[i\sqrt{3}]/(3)$? That would be smaller than $\mathbb{Z}_3[i\sqrt{3}]$ because $3\in(i\sqrt{3})$? If so, what would be such a ring?
Would there be more straight-forward approach?

I would not say that it is "smaller" than  $\mathbb{Z}_3[\sqrt{3}i]$, the question that remains rather is what is $\mathbb{Z}_3[\sqrt{3}i]$  exactly?
We adjoin an element $\alpha$ such that $\alpha^2 = -3$, but this is $0$ in $\mathbb{Z}_3$. This structure is thus formed by  elements $a + b \alpha$ with $a,b \in \mathbb{Z}_3$. Thus with additive structure only  it is $\mathbb{Z}_3^2$. 
Since $\alpha^2=0$, it cannot be a field as it has zero-divisors. Thus it is isomorphic to $\mathbb{Z}_3 \times \mathbb{Z}_3$ with coordinatewise operations as a ring too. 
A possibly somewhat cleaner and in any case more common way to see this is as proposed in comments the other way round first observe the isomorphism to $\mathbb{Z}_3[x]/(x^2+3)$. Yet then $x^2+3 = x^2$ and we are essentially in the same situation as before. (Note we take a quotient with respect to a polynomial that is not irreducible over $\mathbb{Z}_3$ thus the result is not a domain.)
