$I=(3,x^2+1)$ be an ideal of $\mathbb{Z}[x]$. Then $I$ is a proper ideal of $\mathbb{Z}[x]$ This is from an old exam.
Here is the problem.
Let $I=(3,x^2+1)$ be an ideal of $\mathbb{Z}[x]$. Then $I$ is a proper ideal of $Z[x]$
How does one go about showing that this is a proper ideal. Should I consider $Z[x]/I$ and then show that it is neither $Z[x]$ nor $0$ thus showing that $I$ must be a proper ideal or is there some other easier way to proceed?
My attempt: Since $I=\{ 3p(x)+(x^2+1)q(x): p(x), q(x) \in Z[x] \} $ we know that the constant term of $f(0)$ of any element $f(x) \in I$ must be of the form $3n+m$ for $m,n \in \mathbb{Z}$. Does this help me in anyway?. Am I making this too complicated than it should be?
Can you kindly help?
Thank you in advance for your answers.

How far are you with the ring isomorphism theorems to show that $\mathbb{Z}[X]/(3, X^2+1) \cong \mathbb{F}_{3^2} $? Hint: $X^2+1$ is irreducible over $\mathbb{F}_{3}$.
