Conjecture If $(s,t)$ and $(u,v)=(s+i,t+j)$ are in $\{{(x,y) \mid |x^2-dy^2| \le B\}}$ then, under the right conditions, $$i \ge \frac{s}{2B}-\frac{s}{8B^2}.$$ (Near) equality occurs when $s^2-dt^2=\mp B,$ $u^2-dv^2=\pm B$ and $i^2-dj^2=\pm 1.$
By the right conditions I mean something vaguely like $d\sqrt{d} \lt B$ and $B\sqrt{B}<s.$ Also, $i \gt \frac{s}{2B}$ is probably accurate enough except, perhaps, for small $B.$ This is based on rough calculations combined with limited numerical data. I'd suggest clarifying the question, gathering more numerical data and improving the conjecture before attempting a more careful justification. However I think the following observations will help clarify the reasoning. I'll also attempt to relate it to the question as asked in terms of $T$ and $k.$. I don't really understand those conditions but I mainly looked at them before $B$ was corrected to $\sqrt{B}.$
If $x^2-dy^2=C$ (so here $-B \le C \le B$) then $x-y\sqrt{d} \approx \frac{C}{2x}$ and $x+y\sqrt{d}\approx 2x-\frac{C}{2x}.$ I'll assume that $x,y \ge 0$ so we are looking at the set $$S_{d,B}=\{{(x,y) \in \mathbb{N}^2\mid |x^2-dy^2|\le B\}}$$ of lattice points (in the positive quadrant) between the two assymptotes $y=\frac{x}{\sqrt{d}}\pm \frac1{2\sqrt{d}x}$ to the line $y=\frac{x}{\sqrt{d}}.$ Specifically the minimum gap between points $(s,t)$ and $(u,v)=(s+i,t+j)$ in that set. I'll measure the gap as simply $u-s=i$ since $v-t=\frac{i}{\sqrt{d}}+O({\frac{1}{s^3}}).$
So when $s^2-dt^2=-B,$ $u^2-dv^2=B$ and $i^2-dj^2=-1$ we can take the equations
$t\sqrt{d}=s+\frac{B}{2s}$
$v\sqrt{d}=u-\frac{B}{2s}$
$j\sqrt{d}=i-\frac{1}{2i}$
$(s,t)+(i,j)=(u,v)$
Combining these leads to $$B-{\frac {s}{i}}-{\frac {{B}^{2}}{4{s}^{2}}}+{\frac {B\,i}{s}}-1-\frac{1}{4{i}^{2}}
=-B.$$ Dropping the very small terms leaves
$$2B-{\frac {s}{i}}+{\frac {B\,i}{s}}-1=0.$$
Solving this quadratic equation for $i$ (and using $\sqrt{4B^2+1} \approx 2B+\frac{1}{4B}$) yields
$$i=\frac{s}{2B}-\frac{s}{8B^2}.$$
DATA: For $d=2$ and $B=500$ there are $245$ points in $S_{2,500}$ with $10^6 \lt x \lt 2\cdot 10^6.$ The differences $(i,j)$ between successive points come out to $$[[47321, 33461], 139], [[66922, 47321], 62], [[114243, 80782], 27], [[19601, 13860], 16]$$ Here $i^2-2j^2$ is $1,2,-1,1$ respectively. Also, $1.028\frac{s}{2B} \lt i \lt 5.54\frac{s}{2B}$
The lower bound on $i$ is achieved for the points $$(s,t)=(19061766, 13478704),(u,v)=(19081367, 13492564) $$ with $$(i,j)=(19601, 13860).$$ Note that $s^2-2t^2=-476,$ $u^2-2v^2=497$ and $i^2-2j^2=1.$
Since we have two points very close to the line, It seems reasonable to expect $|i^2-dj^2|<B.$ Perhaps even $|i^2-dj^2|<d$ (provided things are appropriately sized.) The best case for a (relatively) small gap would seem to be if $$s^2-dt^2=\mp B,\ i^2-dj^2=\pm 1,\ u^2-dv^2=\pm B.$$ This leads to the computations above.
Aside: I'm not sure if it is needed, but it would be reasonable to conjecture that the number of points in $S_{B,d}$ with $a\lt x \lt b$ is roughly $$\int_a^b\frac{B}{x\sqrt{d}}=\frac{B}{\sqrt{d}}\ln(\frac{b}{a}).$$ That is indeed quite accurate, where I looked, but it fails badly for $B=1$ where we know that the solutions to $x^2-dy^2=\pm1$ are given by $x+y\sqrt{d}=(x_0+y_0\sqrt{d})^i$ for $(x_0,y_0)$ the smallest positive solution. For $d=61$ that is $29718+3805\sqrt{61}$ while it is $8+1\sqrt{63}$ for $d=63.$ In both cases it fails badly. However the estimate was quite accurate in both cases (and also for $d=2$) for $B=500.$
Of course the $x$ values which occur are from a few geometric progressions (with gentle rounding) all with ratio $x_0+y_0\sqrt{d}$ coming from specific values of $x^2-dy^2$ however they appear to blend together well if $B$ is not too small.
For example $|x^2-2y^2|$ can take the values $50,56,62$ but none in between. So $B=50$ and $B=55$ are exactly the same and $B=62$ may not be that different from $B=50.$
Let me now turn to the question as asked. Something looks wrong here but perhaps someone else will want to fix it.
Rather than starting with $T,$ first pick an integer $k$ and then pick $T$ with $k\sqrt{B} \lt T \lt 2k\sqrt{B}.$ Using only the lower bound, the condition $s+t\sqrt{d} \gt T\sqrt{B}$ essentially becomes $$s \gt \frac{Bk+\sqrt{B^2k^2 + 4C}}{4}$$ Where $C=s^2-dt^2$ so $-B \le C \le B.$
It seems pretty safe to use $ \sqrt{B^2k^2 + 4C} \approx Bk+\frac{2C}{kB}$ if $k$ is not too small,but maybe not when $k=1,|C|=B.$
Forging ahead, $$s \gt \frac{Bk}{2}+\frac{C}{2kB}.$$ The other condition $|s-t\sqrt{d}| \lt \sqrt{B}T^{-1}$ becomes $\frac{|C|}{2s} \lt \frac1{k}$ so $$\frac{|C|k}{2} \lt s \lt |C|k.$$
If that seems reasonable, then the bound above of $i \ge \frac{s}{2B}$ becomes $$i \gt \frac{k}{4}+\frac{C}{4B^2k}.$$ That does not seem too promising since $i$ is an integer. Is that really the correct bound for what you want?
