Cardinality of the quotient ring $\mathbb{Z}[x]/(x^2-3,2x+4)$
Your start is good as long as you can show that $2\in(x^2-3,2x+4)$. Fortunately this is the case: $x(2x+4)-2(x^2-3)=4x+6$ and $2(2x+4)-(4x+6)=2$. (Actually one can show that $(x^2-3,2x+4)=(x^2+1,2)$.) So $$\mathbb{Z}[x]/(x^2-3,2x+4)\simeq \mathbb{Z}_2[x]/(x^2+1)$$ and this ring has $4$ elements.
A slight modification of YACP's answer:
If a ring contains an element $x$ satisfying $x^2=3$ and $2x=-4$, then we get $6=2x^2=-4x=8$, hence $2=0$, and the second relation becomes superflous. This shows that $\mathbb{Z}[x]/(x^2-3,2x+4)=\mathbb{Z}/2[x]/(x^2-3)$.
I just want to emphasize that you don't have to come up with clever linear combinations of the generators of the ideal. Just compute inside the quotient ring!
(In fact, as usual, you can also see this as an application of the Yoneda Lemma: The algebraic manipulations above show that $\hom(\mathbb{Z}[x]/(x^2-3,2x+4),-) \cong \hom(\mathbb{Z}/2[x]/(x^2-3),-)$.)
Hint $\ $ Note $\,(2\!-\!x)(2\!+\!x) = (2\!-\!\sqrt 3)(2\!+\sqrt 3)=1\,$ in $\ \Bbb Z[x]/(\color{#0a0}{x^2\!-\!3})\cong \,\Bbb Z[\sqrt 3],\, $ therefore, since $\ 2\!+\!x\,$ in a unit $\!\bmod \color{#0a0}{x^2\!-\!3},\,$ we can cancel it from the generator $\,\color{#c00}2(2\!+\!x)\, $ of $I,\,$ yielding
$\qquad\ \ I = (\color{#0a0}{x^2\!-\!3},\,\color{#c00}2(2\!+\!x)) = (x^2\!-\!3,\color{#c00}2) = ((x\!+\!1)^2,2),\ $ so $\,\ \Bbb Z[x]/I = \Bbb Z_2[x]/((x\!+\!1)^2)$
Remark $\ $ Alternatively we can do the cancellation purely equationally as below
$\bmod I\!:\,\ \color{#0a0}{x^2\equiv 3},\,\ \color{}2(\color{}{2\!+\!x)\equiv 0} \,\Rightarrow\,0\equiv \color{}2(2\!+\!x)(2\!-\!x) \equiv 2(4\!-\!\color{#0a0}{x^2})\equiv 2(4\!-\!\color{#0a0}3) \equiv \color{#c00}2\ $
The resulting quotient ring is $\,\cong \Bbb Z[\epsilon]/(\epsilon^2),\,$ the ring of dual numbers over $\,\Bbb Z_2.\,$ Such dual number rings prove useful for algebraically modelling derivatives, tangent spaces, etc.
The hint amounts to Euclid's Lemma in ideal form, i.e.
Lemma $\,\ \color{#0a0}{(a,b)=1}\,\Rightarrow\, (a,bc) = (a,c)\ $
Proof $\ \ \ (a,bc) = (a,ac,bc) = (a,\color{#0a0}{(a,b)}c) = (a,c)$