Solving an ideal opamp-circuit leads to a contradiction
The error is in the assumption the I2 = I1.
The OpAmp can (in general) sink and source current.
When it would be sourcing current, this current would have to go to ground, either through R2 or through R1 (less likely).  

The current through R2 returns to Vx as well as to the negative power terminal of the source that sources the OpAmp.



Note you cannot omit the power terminals: an ideal OpAmp has no current entering its input terminals. If a current would enter or leave the output terminal, this would conflict with KCL: current cannot be created or eaten by an OpAmp.