Find transfer function of Op Amp system
Your assumption that \$V^- = V^+ \$ is correct in order to avoid the operational amplifier to saturate, and you are right in affirming that \$V_{out}=IR_2\$, where \$I\$ is the current flowing across the \$C\$-\$R_1\$ series as well as across \$R_2\$ due to the fact that no current flows into an ideal op amp (\$I^-=I^+=0\$), so that \$V^+=I^+R_3=0=V^-\$.

Knowing what \$V^-\$ is, you just need to compute \$I\$ from \$V_{in}\$ and the \$C\$-\$R_1\$ series, then substituting it into the expression \$V_{out}=IR_2\$ in order to obtain the correct transfer function.

As you may have realized, your mistake was the computation of the current through voltage divider across \$V^-\$, which is 0 and thus doesn't allow such an operation.