Let $ S $ be the sum of all of Brian's test scores up to this point, and let $ n $ be the number of tests Brian has taken up to this point. Thus the arithmetic mean of his scores now is $ \frac{S}{n} $ and the arithmetic mean of his scores after getting a 98 on the last test will be $ \frac{S+98}{n+1} $. This gives the system of equations: \begin{align*} \frac{S}{n} &= 91 & \frac{S+98}{n+1} & = 92 \end{align*} From the first equation we have $ S = 91n $. Substituting this into the second equation gives: \begin{align*} \frac{S+98}{n+1} &= 92\\ S+98 &= 92(n+1)\\ 91n+98 &= 92n+92\\ 92n-91n&= 98-92\\ n&= 6 \end{align*} So Brian has to take $ n+1 = \boxed{7} $ tests.