the only such I could find was [t] (obviously, since t^2 - 5 is a multiple of t^2 - 5). Here's my reasoning for suspecting there are no others:
suppose [a+bt]^2 =  for some a,b in Q. then we have
(a+bt)^2 - 5 = q(t^2 - 5) for some polynomial q, and rewriting:
a^2 + 2abt + (b^2)(t^2) = qt^2 + 5(q-1)
since the poly on the left is of second degree, so is the poly on the right... hence q must be of degree zero (a number), lest we have a t^3 term. Further, suppose that a is non-zero. Since the right has no t term, we must have 2ab = 0, and hence b = 0, and thus q = 0. then we are left with a^2 = -5, and there is certainly no such a. Thus a must be zero. so we are left with
b^2 t^2 = qt^2 + 5(q-1)
but now 5(q-1) must vanish, and so q = 1, and b=1. This was our original choice.
a zero divisor is by definition a non-zero element which can be multiplied with another non-zero element to yield zero. if f and g are such that
fg = q(t^2 - 5)
for some q, then since t^2 - 5 is irreducible, it must be a factor of either f or g, making one of their equivalence classes equal to  = [t^2-5]. So there are no zero divisors.
The equivalence class [a+bt] has an inverse precisely so long as a and b are not both zero. I'll leave out the calculation, but given any a and b not both zero, let
c = -a/(5b^2 - a^2) d = b/(5b^2 - a^2)
and you can show that (a + bt)(c + dt) is in the equivalence class of 1.
Hope that helps, don't crucify me if there are some typos, I did this late at night. Look closely at c), if your teacher says there's another one there very well may be, but I'd like to know where my reasoning is flawed. take care