2N7000 level shifter doesn't work under a certain voltage
The 2N7000 datasheet I just looked at shows a \$V_{GS(th)}\$ with a minimum of 0.8V and a maximum of 3.0V. When \$V_{TX1.8} = 0V\$, I'd imagine you'd want \$V_{GS} \ge 3.0V\$ so that you can expect the 2N7000 to be ON and pulling down on the 10k\$\Omega\$ output pull-up and presenting about 0V at the \$V_{TX3.4}\$ side of things. Then when \$V_{TX1.8} = 1.8V\$, I'd imagine you'd want \$V_{GS} \le 0.8V\$ so that you can expect the 2N7000 to be OFF and allowing the 10k\$\Omega\$ output pull-up to do its job and present about 3.4V at the \$V_{TX3.4}\$ side of things. Isn't that about it?

The problem is that you can't easily have both cases here, if you can only set the gate voltage with a resistor divider. You'd need to have \$V_G \ge 3.0V\$ in one case and \$V_G \le 2.6V\$ in the other case. So if I were stuck with the 2N7000 and planning on shooting for a circuit that will probably work okay, I'd probably give up and set \$V_G \approx 2.5V\$ and just go with that.

This brings up another thing. You've got a 10k\$\Omega\$ resistor going from the resistor divider node back to the \$V_{TX1.8}\$ node (source.) Why? This actually moves the gate closer to ground at a time when you'd like it further away, and visa versa. I think I'd dump that. It could just be that I'm ignorant (I'm just a hobbyist and have no formal training at all in electronics.) But it doesn't make any sense to me, right now. It actually seems to push things the wrong way -- to me, anyway.

So perhaps try something like this?







simulate this circuit &ndash; Schematic created using CircuitLab

Now:

\$V_{TX1.8} = 0V \rightarrow V_{GS} = 2.45V\$, transistor is \$\approx\$ ON

\$V_{TX1.8} = 1.8V \rightarrow V_{GS} = 0.65V\$, transistor is \$\approx\$ OFF

Which I think should do about what you want.