You can see from this:
That if the drain and gate are tied together and the drain current is modest, then at most you need \$V_{_\text{GS}}\le 4\:\text{V}\$. There's also a transfer curve, later:
That typically at \$V_{_\text{GS}}= 5\:\text{V}\$ the drain current can be quite high regardless of operating temperature. Certainly, high enough for most LEDs.
You will need to use a method to limit the LED current. If you are excited by the idea of using this particular MOSFET and if your supply voltage for the LED is more than about \$2\:\text{V}\$ (which it is, as you say) more than the LED itself needs, then you can add a BJT like this:
simulate this circuit– Schematic created using CircuitLab
So, if the LED current is \$20\:\text{mA}\$ then \$R_1\approx 10\:\text{k}\Omega\$, \$R_2\approx 22\:\text{k}\Omega\$, and \$R_{_\text{SET}}\approx 33\:\Omega\$.
Feel free, though, to use a different value for \$I_{_\text{LED}}\$ that meets your needs.
\$Q_1\$ can be any junkbox small signal NPN BJT.
(\$R_1\$ and \$R_2\$ are arranged to provide a voltage divider that keeps the NFET gate voltage within a valid/acceptable range.)