$I=(3,x^2+1)$ 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.

Boring way: show that a single element, say 1, is not in your ideal.  Do this by contradiction --- what can you say about p(x) and q(x) if 1 is in $I$?
Fun way:  set $R = \mathbb{Z}[x]/(x^2+1)$, and show that
1)  R is isomorphic to the ring of Gaussian integers $\mathbb{Z}[i]$, and
2) the quotient $\mathbb{Z}[x]/I$ is isomorphic to $\mathbb{Z}[i]/(3)$
Then use your favorite method to show that $3$ is not a unit in $\mathbb{Z}[i]$, so generates a proper ideal.
