(I edited heavily my original post because it was based on my serious misunderstanding of a textbook exercise. However the main question remains valid as explained below.)
It appears to me (and also pointed out in comments below) that the following circuit does not have a Thevenin equivalent.
simulate this circuit – Schematic created using CircuitLab
I take branch directions as follows: \$V_1\$ from 0 to 1, \$R_2\$ from 2 to 1, \$R_3\$ from 0 to 2, \$I_4\$ from 2 to 3, \$R_5\$ from 3 to 4. For clarity, the CCCS is \$I_4=8i_{R_2}\$.
To find the Thevenin equivalent I connect as load a voltage source \$V_L\$ from 4 to 0. To solve the circuit I take loops \$V_1R_2R_3\$, \$R_3I_4R_5V_L\$; call the loop currents \$p_1,p_2\$. In \$V_1R_2R_3\$ we have: $$ V_1+R_2p_1+R_3(p_1-p_2)=0\Rightarrow 5+10p_1+10(p_1-8i_2)=0\Rightarrow 5+10p_1+10(p_1+8p_1)=0\Rightarrow p_1=-0.05. $$ Then $$ i_L=i_5=i_4=8i_2=-8p_1=0.40. $$ In other words, the current through the load \$V_L\$ is \$i_L=0.40\$, independent of \$V_L\$. This cannot be attained by any Thevenin equivalent, because a Thevenin equivalent is a practical voltage source and must have characteristic \$i_{Th}=\frac{V_{Th}}{R_{Th}}\$.
Question 1: So does it follow that the circuit has no Thevenin equivalent?
Question 2: More generally, what conditions must a circuit satisfy to have a Thevenin equivalent?
PS2: The analysis and conclusion remain valid if we use a resistor load \$R_L\$ instead of \$V_L\$. We cannot use a current source load \$I_L\$, because then we would get \$I_L=i_4=0.4\$ (by the same analysis as above). But \$i_L\$ can be arbitrary; it is not obliged to be \$I_L=0.4\$. In other words, the circuit with a current source load has no solution (or an infinity of solutions).
PS2: The circuit does have a Norton equivalent with \$I_{Nr}=0.4\$, \$R_{Nr}=\infty\$. In other words, it seems to me that the original can be replaced by an ideal current source.
