Prove/disprove: $ I_1 \cong I_2 \iff R/I_1 \cong R/I_2 $

No, this is not the case.

Consider, for instance, $R = \Bbb Z[x_1, x_2, x_3, \ldots]$, and the two isomorphic ideals $I_1 = (x_1, x_2, x_3, \ldots)$ and $I_2 = (x_2, x_4, x_8, \ldots)$. Then $R/I_1\cong \Bbb Z$, while $R/I_2\cong R$.

$\mathbb{Z} \times \mathbb{Z} \times \mathbb{Z}^\mathbb{N}$ with ideals $\{0\} \times \mathbb{Z} \times \mathbb{Z}^\mathbb{N}$ and $\{0\} \times \{0\} \times \mathbb{Z}^\mathbb{N}$.

Consider $R=\mathbb{Q}[x_i\ : \ i\in \mathbb{N}]$, $I_1= (x_1, x_2, \dots)$ and $I_2=(x_2, x_3, \dots)$. Those are isomorphic via $x_i \mapsto x_{i+1}$. However, $I_1$ is maximal, wheras $I_2$ is not. Thus the quotients are not isomorphic (the first quotient will be a field, and the second not).