A 555 timer already has a fairly capable 2-quadrant output. One that is much better than your single bipolar with a resistor for its output. Partly for that reason, it wasn't clear to me where you are headed until you wrote in comments:
I am looking to get the same square signal but boosted to somethinglike 15v to open the gate of a MOSFET fully. the 9volt VCC can beincreased, up to 15volts.
That helped a lot.
If you don't mind that the output is inverted from the input, then the following behavioral circuit provides 2-quadrant output which can run easily at \$200\:\text{kHz}\$ with commonly found bipolars and will allow you to set \$V_{_\text{PP}}=15\:\text{V}\$ while keeping \$V_{_\text{CC}}=9\:\text{V}\$(and lower, too, if this is a small battery that gradually declines in output voltage.)
simulate this circuit– Schematic created using CircuitLab
Michal Podmanický's answer reminded me about the speed-ups. So this circuit includes them for both the high and low sides (\$R_7+C_1\$ on one side and \$R_8+C_2\$ on the other.)
\$Q_3\$ operates in (to my mind) cascode mode. \$Q_1\$ and \$Q_2\$ provide the 2-quadrant output drive.
\$R_1\$ and \$R_2\$ provide some modest management of short-circuit/shoot-through and emitter degeneration. (You may remove them and just use a wire if it works for you.)
Here's the part values for \$V_{_\text{PP}}=15\:\text{V}\$ and \$V_{_\text{CC}}=9\:\text{V}\$ with an output compliance of \$100\:\text{mA}\$:
Here's the LTspice run for it, using two runs: (1) essentially unloaded; and (2) fully loaded:
I just picked the usual bipolars. Dissipation, even with some shoot-through, should be under \$200\:\text{mW}\$. Tolerable.
It appears to work reasonably well at \$200\:\text{kHz}\$.