ICIS opamp circuit - current gain derivation
You're not doing anything wrong at all, it may just be that you still need to solve for \$i_{out}\$ from \$V_{out}\$ and \$V_x\$, which could be a bit more tedious.

I suggest adding the variable \$i_{out}\$ and adding an extra equation. The equations are very similar, but it allows you to solve for \$i_{out}\$ directly.

$$-i_{in} + \frac{V_- - V_x}{R_2} = 0$$

$$V_{out} = -A_{VOL}V_-$$

$$\frac{V_x - V_-}{R_2} - i_{out} + \frac{V_x}{R_1} = 0$$

$$\frac{V_{out} - V_x}{R_L} = i_{out}$$

Solving this for \$i_{out}\$ (I did this with a CAS):

$$
\frac{i_{out}}{i_{int}} = -\frac{A_{VOL}R_2+(1+A_{VOL})R_1}{R_L + (1+A_{VOL})R_1}
$$

When assuming that \$A_{VOL} \gg 1\$, you get the expression from your book.

APPENDIX

It is still possible to get the right answer from your equations as well. Solving your equations will give you:

$$V_x = -i_{in}\frac{R_1(A_{VOL}R_2-R_L)}{R_L+(1+A_{VOL})R_1}$$

$$V_{out} = -i_{in}\frac{A_{VOL}R_2(R_L+R_1)+A_{VOL}R_1R_L}{R_L+(1+A_{VOL})R_1}$$

$$i_{out} = \frac{V_{out}-V_x}{R_L} = -i_{in}\frac{A_{VOL}R_2+(1+A_{VOL})R_1}{R_L+(1+A_{VOL})R_1}$$

This is identical.