[Fzang]

Two pairs of pi electrons from the ring itself, a pair from oxygen and a pair from nitrogen. That makes 4 pairs, which shouldn't be aromatic.

But why am I reading "aromatic" wherever I look?

[MissPhosgene]

It is aromatic. 3 pi electrons from the carbons in the ring. Two from oxygen, one from nitrogen.

[Hybrid]

2 pi electron " c=c" & c=n" lone pair of N in sp2 orbital not p orbital + lone pair of O in p orbital. similar to imidazole see here

[Fzang]

So nitrogen only provides an available lonepair if it has 3 bonds?

[Hybrid]

as it "sp2 N" it has 3 sp2 and 1 p so if it make = it will consume the 1 p so every bond after that and its lone pair will be sp2 "not parallel to ring = " so it will be basic.

[cupid.callin]

Acc. to Huckel's rules, a molicule should be planar to be aromatic.
As you asked, N should have 3 bonds, it shoud have a pi bons. if the pi bond C=N is made C-N, it wont be aromatic anymore, you can check by making resonating structures.

If bond b/w C and N is single, due to tetrahedral shape, molicule wont be planar anymore.

[Telamond]

Just want to mention this (since the op's question is already answered); a molecule doesn't have to be planar to be aromatic.
Barrelene is an example of this.

[jake.n]

The hybridization arguments posted above are good.  Typically, if a molecule can be hybridized such that it is aromatic, it will be.  Aromaticity is a strong stabilizing force, and molecules tend to arrange themselves to minimize their energy.