Edit: Made a mistake in the MLE part There is no MLE because the likelihood is maximized at at a parameter specification outside of the parameter range.
\begin{equation} f_T (t;B,C) = \frac{{e^{ - \frac{t}{C}} - e^{ -\frac{t}{B}} }}{{C - B}} \end{equation}
Taking the gradient we get:
\begin{equation} \nabla f_T (t;B,C) = \{\frac{e^{\frac{-1}{x}}}{x^2(x-y)}-\frac{e^{\frac{-1}{x}}-e^{\frac{-1}{y}}}{(x-y)^2},\frac{e^{\frac{-1}{x}}-e^{\frac{-1}{y}}}{(x-y)^2}-\frac{e^{\frac{-1}{y}}}{y^2(x-y)}\} \end{equation}
Setting this equal to 0 and re-arranging some terms gives us the following system of equations:
\begin{array}{lcl} \frac{e^{\frac{-1}{C}}}{C^2} & = & \frac{e^{\frac{-1}{C}}-e^{\frac{-1}{B}}}{(C-B)}\\ \frac{e^{\frac{-1}{B}}}{B^2} & = & \frac{e^{\frac{-1}{C}}-e^{\frac{-1}{B}}}{(C-B)} \end{array}
\begin{array}{lcl} \frac{(C-B)}{C^2} & = & 1-e^{\frac{-1}{B}+\frac{1}{C}}\\ \frac{(C-B)}{B^2} & = & 1-e^{\frac{-1}{C}+\frac{1}{B}} \end{array}
Which is maximized when $C = B$, which is not a valid parameterization.
If you have that $E[X] = C+B$ and $E[X^2] = 2(B^2 + BC + C^2)$ then: \begin{equation} Var[X] = 2B^2 + 2BC + 2C^2 - B^2 -2BC - C^2 = B^2 + C^2 \end{equation}
Using method of moments we get that $C = \bar{X} - B \Rightarrow \hat{\sigma^2} = B^2 + B^2 + \bar{X}^2 -2\bar{X}B \Rightarrow 0 = 2B^2 -2\bar{X}B + \bar{X}^2 - \hat{\sigma^2}$.
The roots of that quadratic get you $B$ (one is probably negative so take the positive one), and plugging that estimate for $B$ into $C = \bar{X} - B$ gives you the moment estimator for C.
Edit: Examining the methods of moment estimators The root for the polynomial in $B$ is $\hat{B} = \frac{1}{2}(\sqrt{2\hat{\sigma}^2 - \bar{X}^2}+\bar{X})$. Plugging that into the expression for $C$ gives $\hat{C} = \frac{1}{2}(\bar{X} - \sqrt{2\hat{\sigma}^2 - \bar{X}^2})$.
Note that with this estimate $\hat{C} < \hat{B}$, but because in the estimating equations we didn't imply an ordering on $B$ or $C$, and because the formula for the mean and variance are symmetric in the parameters, we can just switch the estimators to make $\hat{B} = \frac{1}{2}(\bar{X} - \sqrt{2\hat{\sigma}^2 - \bar{X}^2})$ and $\hat{C} = \frac{1}{2}(\sqrt{2\hat{\sigma}^2 - \bar{X}^2}+\bar{X})$.
Noting that the MLE didn't converge to anything it might be interesting to examine whether these estimates are unbiased for the mean and variance and have asymptotic consistency. \begin{equation} E[\hat{C} + \hat{B}] = E[\frac{\bar{X} + \bar{X}}{2}] = E[\bar{X}] = \mu \end{equation} \begin{equation} \begin{split} E[\hat{C}^2 + \hat{B}^2]& = E[\frac{1}{4}((\bar{X}^2 +2\bar{X}\sqrt{2\hat{\sigma}^2 -\bar{X}^2} + 2\hat{\sigma}^2 - \bar{X}^2)) + (\bar{X}^2 -2\bar{X}\sqrt{2\hat{\sigma}^2 -\bar{X}^2} + 2\hat{\sigma}^2 - \bar{X}^2)] \\ & = E[\frac{2\hat{\sigma}^2 + 2\hat{\sigma}^2}{4}] \\ & = E[\hat{\sigma}^2] = \sigma^2 \end{split} \end{equation}
To show convergence, note that by the law of large numbers convergence in probability to a constant $\bar{X}\overset{P}{\rightarrow}\mu$. Applying Slutsky's theorem to $2\hat{\sigma^2} - \bar{X}^2 \overset{P}{\rightarrow} 2\sigma^{2^*} - \mu^2$, where $\sigma^{2^*}$ is the limiting distribution of $\hat{\sigma}^2$.
Applying the continuous mapping theorem to the square root, then Slutsky's theorem again to the additional $\bar{X}$, we have that:
\begin{equation} \hat{B} \overset{P}{\rightarrow} \frac{1}{2}(\mu- \sqrt{2\sigma^{2^*} - \mu^2}) \end{equation}
\begin{equation} \hat{C} \overset{P}{\rightarrow} \frac{1}{2}(\sqrt{2\sigma^{2^*} - \mu^2} + \mu) \end{equation}
By the law of large numbers $\sigma^{2^*} \overset{P}{\rightarrow} \sigma^2$. Plugging $\sigma^2 = B^2 + C^2$ and $\mu = B + C$ into these expressions gives us:
\begin{equation} \begin{split} \hat{B} & \overset{P}{\rightarrow} \frac{1}{2}(\mu- \sqrt{2\sigma^{2^*} - \mu^2})\\ & = \frac{1}{2}(B + C- \sqrt{B^2 -2BC + C^2})\\ & = \frac{1}{2}(B + C- \sqrt{(C-B)^2})\\ & = \frac{1}{2}(B + C- C + B)\\ & = B \end{split} \end{equation}
\begin{equation} \begin{split} \hat{C} & \overset{P}{\rightarrow} \frac{1}{2}( \sqrt{2\sigma^{2^*} - \mu^2} + \mu)\\ & = \frac{1}{2}(B + C + \sqrt{B^2 -2BC + C^2})\\ & = \frac{1}{2}(B + C + \sqrt{(C-B)^2})\\ & = \frac{1}{2}(B + C + C - B)\\ & = C \end{split} \end{equation}