Can the plane of symmetry for a compound be the plane of the paper?
In short: The presence of the $\sigma_h$ plane you see depends on the conformation of the methyl groups.

If your example were either benzene or 1,4-dichlorobenzene, then you were be right, the molecule would contain a plane of symmetry identical to the plane defined by the benzene ring.  This $\sigma_h$ plane then could be described as in the following example by the yellow disk:



(credit to symmetry@otterbein, example symmetry $C_{6h}$, hexakis(Me2N)benzene).

Your example however is about a methyl-substituted benzene. While the carbon atoms of the methyl groups are in the plane of the symmetry, you no longer can apply the same operation (reflection on this mirror plane) on the hydrogens of these methyl groups to yield corresponding, symmetry equivalent hydrogens on the other side of the mirror.  This is only possible if both the methyl groups share a special conformation: for each exactly one hydrogen must reside in this $\sigma_h$ mirror plane; then, the other two hydrogens of the methyl group in question are symmetry equivalent.

In all other conformations, the three hydrogens of one methyl group were no longer symmetry equivalent in respect to this $\sigma_h$ plane.  And it would require only one methyl group to be «off» to erase the $\sigma_h$ plane for the whole molecule.

I would suggest to venture out the interactive tutorials around symmetry@otterbein.  All you need is an internet browser with javascript enabled.  Later, if interested in applying the concepts of symmetry into crystallography (including the transition of Schoenflies notation used in IR-spectroscopy by the  Hermann-Maguin about point groups), you may find its sibling crystals@otterbein a valuable complement.