Gate Source Voltage Drop Approximation
BJT and MOSFET have different working mechanism. But there are similar aspects indeed. As you see, no matter MOSFET current mirror or BJT mirror, the two transistor have same gate-source (or base-emitter) voltage, but because BJT and MOSFET have different working mechanism, so the (Vcc-0.7/R)  approximation won't work for MOSFET current mirror. 

But this won't prevent you to find a 'rule' to simplify the analysis: 

When short the drain and gate of M1, it will forcing it to operate in the saturation mode, because \$ V_{GS} - V_{DS} = 0 &lt; V_{tn} \$. So, we have
$$
I_{M1} = I_{REF} = \frac{V_{DD}-V_{GS}}{R}\qquad(1)\\
I_{M1} = \frac{1}{2}k_{n}'(\frac{W}{L})_{1}(V_{GS}-V_{tn})^\qquad(2)\\
I_{OUT}=I_{D2}=\frac{1}{2}k_{n}'(\frac{W}{L})_{2}(V_{GS}-V_{tn})^2\qquad(3)\\
$$

From equations (2) and (3), we get

$$
\frac{I_{O}}{I_{REF}}=\frac{(W/L)_2}{(W/L)_1}\qquad(4)
$$

So, Equations (1) and (4) are the rules for MOSFET current mirror. They are easy to remember and use.