Identify $\mathbb{Z}[x]/(2x^2+1,2x-3)$.

That's good. A way to then work backwards is to show that:

$$2x-3,2x^2+1\in\langle 11,x-7\rangle$$

You get: $$\begin{align}2x-3&=2(x-7)+11\cdot1\\ 2x^2+1&=2(x-7)(x+7)+11\cdot 9 \end{align}$$