I know the method of continued fractions to solve the Pell's equation. I need help turning $2b(b-1) = t(t-1)$, with $b, t$ as integers, into the form $x^2 - ny^2 = 1$, if possible.
My attempt is $(t-1/2)^2 - 2(b - 1/2)^2 = -1/4$, but those aren't integer solutions anymore.
Your completing the square procedure is good. Multiply through by $4$. We get $(2t-1)^2-2(2b-1)^2=-1$. Not quite the Pell equation, but a close relative, $a^2-2b^2=-1$. The fundamental solution is $a=1$, $b=1$, and all positive solutions can be obtained as for the Pell equation. One way to generate them is that they are $(a_k,b_k)$, where $(1+\sqrt{2})^{2k+1}=a_k+b_k\sqrt{2}$.
Alternately, one can get a recurrence for the "next" solution $(a_{k+1},b_{k+1})$ in terms of the current solution $(a_k,b_k)$.
Added: For reference, the recurrence turns out to be $a_{k+1}=3a_k+4b_k$, $b_{k+1}=2a_k+3b_k$. From this it is easily checked that all the solutions are odd, so they all generate a solution of your original equation. The first few solutions $(a,b)$ of the equation $a^2-2b^2=-1$ are: $$(1,1),\quad (7,5),\quad (41,29),\quad (239,169),\quad (1393,985).$$