You can solve this circuit by *more or less* the same method you've given in the question; however you need to plug in one more equation (\$V_1=V_2+12\$) into the system and introduce an unknown current variable. So I'm not sure if we can call it a 'pure' nodal analysis.

This is what you've got to do:

* Write KCL on the left node (\$I_s\$ is the current through the voltage source): $$6A=\frac { { V }_{ 1 }-{ V }_{ 2 } }{ { R }_{ 3 } } +\frac { { V }_{ 1 } }{ { R }_{ 3 } } +{ I }_{ s }$$

* Do it again on the node on the right side:
$$4A={ I }_{ s }-\frac { { V }_{ 2 } }{ { R }_{ 2 } } +\frac { { V }_{ 1 }-{ V }_{ 2 } }{ { R }_{ 3 } } $$

* So far, we're in line with the method described in the question. As a last step, write down this one:
$$V_1=V_2+12$$

Now we're left with 3 equations in three unknowns, which you can easily solve to get \$V_1, V_2 and I_s\$