If you know the correspondence theorem this is immediate (i.e. if R/I has a non-trivial proper ideal J then consider the corresponding ideal J' in R. The ideal J' contains I so must be either R or I by maximality of I. The first contradicts the fact that J was proper, the second contradicts non-triviality)
I am assuming therefore, that you don't know and/or aren't expected to know the correspondence theorem. In that case; it is a bit harder to think up.
Hint: For each non-zero element x+I in R/I and consider the ideal J = I + Rx.
Since x isn't in I (else x+I would be zero in R/I), I is strictly contained in J, whence by maximality of I we have that J = R. Thus in particular - the identity element 1 of R is in J and so we may write 1 = i + rx for some r in R, i in I. It then follows that (x+I)(r+I) = xr +I = xr + i + I = 1 + I so that (x+I) is invertible as required.
Last edited by Jake22; 11-05-2012 at 15:16.