A knight can't make a tour on a $2\times n$ board, for any $n$,
because it must go into and out of a corner square, and it can't do
this without going back on itself.
On the $3\times 4$ grid, we must use a path from the loop JAGIBHJ
and a path from the loop KDFLCEK. But they only link up between J
and C, and between B and K. So the path must start at a neighbour
of J, B, K or C, follow round that loop, switch to the other loop
and follow round that. Obviously the path can go round the loop in
either direction. So there are $16$ possible tours:
HJAGIBKDFLCE
HJAGIBKECLFD
HBIGAJCEKDFL
HBIGAJCLFDKE
AGIBHJCEKDFL
AGIBHJCLFDKE
IGAJHBKECLFD
IGAJHBKDFLCE
ECLFDKBIGAJH
ECLFDKBHJAGI
EKDFLCJHBIGA
EKDFLCJAGIBH
DFLCEKBIGAJH
DFLCEKBHJAGI
LFDKECJAGIBH
LFDKECJHBIGA
Since it's not possible to get from the finish directly back to the
start in any of these tours, there is no circuit.