if $d\equiv 1\pmod{3},d>1$,then $3\mid d^2+d+1,d^2+d+1>3$ if $d\equiv 2\pmod{3},d>2$,then $3\mid d^2-d+1,d^2-d+1>3$ so $n=2\cdot 3^k$ or $3^k$ Notice that $9^2+9+1=91=7\times 13$ $k<2$,$n=2,3,6$