Solving for current with voltage drops
I'll get right to the point here. I don't know the actual parameter values that your simulator is using. But it is probably using something like \$\eta=4\$ (the emission co-efficient.) You also already know about \$V_T=\frac{k\,T}{q}\$ as the thermal temperature voltage. I'm going to assume a saturation current of \$100\:\text{pA}\$ for the LED and thereby move directly to a closed solution.

The diode voltage is:

$$V_\text{D}=\eta\cdot V_T\cdot\operatorname{ln}\left(\frac{I_\text{D}}{I_\text{SAT}}+1\right)$$

Using KVL, we find the following (\$V_\text{CC}\$ is the power supply voltage):

$$V_\text{CC}-I_\text{D}\cdot R-\eta\cdot V_T\cdot\operatorname{ln}\left(\frac{I_\text{D}}{I_\text{SAT}}+1\right)=0\:\text{V}$$

That's basically it. You start with \$V_\text{CC}\$, subtract the voltage drop across the resistor, \$R\$, and then subtract the diode voltage and you should have nothing left -- just \$0\:\text{V}\$.

Let's set \$I_T=\eta\cdot \frac{V_T}{R}\$. Then:

$$I_\text{D}=I_T\cdot \operatorname{LambertW}\left(\frac{I_\text{SAT}}{I_T}\cdot e^{^{\frac{\frac{V_\text{CC}}{R}+I_\text{SAT}}{I_T}}}\right)-I_\text{SAT}$$

That's how you can compute the current in the LED.

Obviously, once you have \$I_\text{D}\$ you can then compute:

$$V_\text{D}=\eta\cdot V_T\cdot\operatorname{ln}\left(\frac{I_\text{D}}{I_\text{SAT}}+1\right)=I_T\cdot R\cdot\operatorname{ln}\left(\frac{I_\text{D}}{I_\text{SAT}}+1\right) $$

If you need to see how I derived the equation for \$I_\text{D}\$, I can provide that derivation. I just felt I'd save you the added learning, for now.