Prime ideal of $\Bbb Z[X]$ is $(0)$ or $(f)$ or $(p, f)$
($f\in\Bbb Z[X]$, $f$ is an irreducible element. And $p$ is a prime number)
By the statement above, $(x^3+2,2x^2+3)$ is not a prime ideal.
I don't know how to check this...
I think there is no way to make $(x^3+2,2x^2+3)$ simpler. (I mean simpler by transforming $(x^2, x+1)$ to $(x+1, 1)$)
$$3x-4 = x(2x^2+3) - 2(x^3+2)$$
$$-8x-9 = 2x(3x-4) - 3(2x^2+3)$$
$$x-21 = 3(3x-4) + (-8x-9)$$
So we have: $$(x^3 + 2, 2x^2 + 3) = (x^3 + 2, 2x^2 + 3, x-21)$$ $$= (21^3 + 2, 2\cdot 21^2 + 3, x-21) = (9263,885,x-21)$$
But $\gcd(9263,885)=59$, so $(x^3 + 2, 2x^2 + 3) = (59,x-21)$. From here it is easy to see that the quotient is $\mathbb{Z}/59$. (Or, you can take $p=59$ and $f=x-21$, and use the first sentence of the question!)