I have the following function where $F(x)$ is a normal distribution and in this context it represents a 'weight': $$R(t) = \int_\tau \int_x T(\tau)\cdot F(x) \cdot \delta(t -(mx+b) - \tau) \: dx \: d\tau$$
I cannot isolate $F(x)$ and integrate it to 1 since the $(mx+b)$ is inside of the dirac delta function. They are interdependent. I originally tried isolating the $F(x)$ and integrated it to 1 then I solved for the following: $$= \int_\tau T(\tau) \cdot \delta(t -(mx+b) - \tau) \: d\tau \int_x \: F(x)dx $$ $$=\int_\tau T(\tau) \cdot \delta(t -(mx+b) - \tau) \: d\tau$$ $$ =T((mx+b) - t) $$
I checked this in Mathematica and got the same result, but I took the wrong steps to get there. What would've been the correct steps instead of trying to isolate $F(x)?$
So what is wrong in your case is only the fact in your second equation, you cannot write the $x$ integral after the Dirac as it is depending on $x$. But you can write for example $$ \iint T(\tau)\,F(x)\,\delta(t-mx-b-\tau)\,\mathrm d x\,\mathrm d τ = \int F(x) \left(\int T(\tau)\,\delta(t-mx-b-\tau)\,\mathrm d τ\right)\mathrm d x \\ = \int F(x)\, T(t-mx-b)\,\mathrm d x $$ using the definition of the Dirac delta as a measure in the $\tau$ variable. This is now an integral that depends on your specific $F$ and $T$. In general, you cannot simplify it much further. If $F$ is a Gaussian, then you can expand the square in the Gaussian to factorize the $t-b$ part.
Remark: the last integral can be written as a convolution. If $T_m(x) = T(mx)$ then $$ \int F(x)\, T(t-mx-b)\,\mathrm d x = \int F(x)\, T_m(\tfrac{t-b}{m}-x)\,\mathrm d x = (F * T_m)(\tfrac{t-b}{m}) $$