This will be a partial answer at this point. Maybe I will come back and say more later.

For now I will assume you know the value of $\sigma_c,$ as opposed to needed to estimate it based on the data. With a sample consisting of six binary observations $I_1,\ldots,I_6$ let us say $$ I_i = \left. \begin{cases} 0 & \text{if } v_1+\text{noise} <v_2+\text{noise} \\ 1 & \text{if } v_1+\text{noise}>v_2+\text{noise} \end{cases} \right\} \text{ for } i=1,2,3,4,5,6. $$ If we express $v_1+\text{noise} <v_2+\text{noise}$ by subtracting $v_2$ from both sides and subtracting the noise on the left side from the right side, we get $$ v_1-v_2 < \text{noise}, $$ where this latest noise term as variance $2\sigma_c^2.$

Then \begin{align} & L(v_1,v_2) \\[8pt] = {} & \prod_{i\,:\,I_i\,=\,0} \Pr(I_i=0 \mid v_1,v_2) \cdot \prod_{i\,:\,I_i\,=\,1} \Pr(I_i=1 \mid v_1,v_2) \\[8pt] = {} & \prod_{i\,:\,I_i\,=\,0} \Pr \big( \text{noise} > v_1-v_2 \big) \cdot \prod_{i\,:\,I_i\,=\,1} \Pr \big( \text{noise} > v_1-v_2 \big) \\ & \text{where this “noise” has variance } 2\sigma_v^2, \\[8pt] = {} & \prod_{i\,:\,I_i\,=\,0} \Pr\left( \frac{\text{noise}}{\sigma_v\sqrt2} > \frac{v_1-v_2}{\sigma_v/\sqrt2} \right) \cdot \prod_{i\,:\,I_i\,=\,1} \Pr\left( \frac{\text{noise}}{\sigma_v\sqrt2} < \frac{v_1-v_2}{\sigma_v\sqrt2} \right) \\[8pt] = {} & \prod_{i\,:\,I_i\,=\,0} \left( 1 - \Phi\left( \frac{v_1-v_2}{\sigma_v\sqrt2} \right) \right) \prod_{i\,:\,I_i\,=\,1} \Phi\left( \frac{v_1-v_2}{\sigma_v\sqrt2} \right). \end{align} If you get at least one $0$ and at least one $1,$ then you can compute the value of $v_1-v_2$ that maximizes this, so you have an MLE for the difference.

If you get all $1\text{s}$ or all $0\text{s},$ the maximizing value of $v_1-v_2$ will be $-\infty$ or $+\infty.$