Thank you Jim, your contributions about statistics are also welcome! The Mathematica message error was explained by Gachevski. It is related to number machine precision, replacing a number with number-dot: 1 -> 1.

I owe you a more substantial demonstration as to why the "regression approach" should be avoided at all costs. Below are the results of a 100 simulations of a random sample of size 3705 (your original sample size) from a Rosin-Rammler distribution with parameters $n=4.5$ and $dm=200$. You'll see that the regression approach is extremely biased for estimating $dm$ and extremely variable for both $n$ and $dm$.

nsim = 100;  (* Number of simulations *)
mle = ConstantArray[{0, 0}, nsim];
reg = ConstantArray[{0, 0}, nsim];
mleBinned = ConstantArray[{0, 0}, nsim];
nSamples = 3705;
Do[
 x = RandomVariate[WeibullDistribution[4.5, 200], nSamples];
 bins = N[PowerRange[1, 14000, 2^(1/3)]];
 dpimean = ConstantArray[0, Length[bins] - 1];
 Do[dpimean[[i]] = (bins[[i]] + bins[[i + 1]])/2, {i, 
   Length[bins] - 1}];
 c = BinCounts[x, {bins}];
 sizedistcum = Transpose[{dpimean, 1. Accumulate[c]/nSamples}];

 (* Regression estimates of n and dm *)
 (* Select only the cumulative proportions strictly between 0 and 1 *)
 data = Select[sizedistcum, 0 < #[[2]] < 1 &];

 (* Transform the coordinates *)
 data[[All, 1]] = Log[data[[All, 1]]];
 data[[All, 2]] = Log[-Log[1 - data[[All, 2]]]];

 (* Fit a line corresponding to a distribution function and show results *)
 lm = LinearModelFit[data, z, z];

 (* Get regression estimates of n and dm *)
 {intercept, nreg} = lm["BestFitParameters"];
 dmreg = Exp[-intercept/nreg];
 reg[[i]] = {nreg, dmreg};

 (* Maximum likelihood estimates with raw data *)
 mle[[i]] = {n, dm} /. FindDistributionParameters[x, WeibullDistribution[n, dm]];

 (* Maximum likelihood estimates with binned data *)
 (* Cumulative distribution function *)
 n =.;
 dm =.;
 F[x_] := 1 - Exp[-(x/dm)^n];

 (* Log of likelihood *)
 logL = Sum[c[[j]] Log[F[bins[[j + 1]]] - F[bins[[j]]]], {j, Length[c]}];

 (* Find maximum likelihood estimates *)
 mleBinned[[i]] = {n, dm} /. (FindMaximum[{logL, n > 0 && dm > 0}, {{n, mle[[i, 1]]}, {dm, mle[[i, 2]]}}, WorkingPrecision -> 30][[2]]),

 {i, nsim}]

(* Show results *)
ListPlot[{mle, mleBinned, reg, {{4.5, 200}}}, Frame -> True, 
 FrameLabel -> (Style[#, Bold, 18, Black] &) /@ {"n", "dm"},
 PlotStyle -> {{PointSize[0.01], Blue}, {PointSize[0.01], 
    Green}, {PointSize[0.01], Orange}, {PointSize[0.025], Red}},
 PlotLegends ->  PointLegend[{Blue, Green, Orange, Red}, {"Maximum likelihood (raw data)", 
    "Maximum likelihood (binned data)",  "Regression on sample distribution function", "True values"},
   LegendMarkerSize -> 20], ImageSize -> 500, PlotRange -> All]

If the raw data isn't available, then one can still obtain maximum likelihood estimates from the binned data (assuming the binned data can be transformed into the observed frequency counts).

sizedistcum = {{1.12996, 0.}, {1.42366, 0.}, {1.7937, 0.}, {2.25992, 
    0.}, {2.84732, 0.}, {3.5874, 0.}, {4.51984, 0.}, {5.69464, 
    0.}, {7.1748, 0.}, {9.03968, 0.}, {11.3893, 0.}, {14.3496, 
    0.}, {18.0794, 0.}, {22.7786, 0.}, {28.6992, 0.}, {36.1587, 
    0.}, {45.5572, 0.}, {57.3984, 0.00215924}, {72.3175, 
    0.0283401}, {91.1143, 0.0537112}, {114.797, 0.0753036}, {144.635, 
    0.103104}, {182.229, 0.268826}, {229.594, 0.832659}, {289.27, 
    0.964372}, {364.457, 0.993522}, {459.187, 1.}, {578.54, 
    1.}, {728.914, 1.}, {918.375, 1.}, {1157.08, 1.}, {1457.83, 
    1.}, {1836.75, 1.}, {2314.16, 1.}, {2915.66, 1.}, {3673.5, 
    1.}, {4628.32, 1.}, {5831.32, 1.}, {7347., 1.}, {9256.64, 
    1.}, {11662.6, 1.}};

(* Frequency counts for each bin *)
c = Differences[Round[3705 sizedistcum[[All, 2]]]]
(* Bin borders *)
x = N[PowerRange[1, 13000, 2^(1/3)]]

(* Cumulative distribution function *)
F[x_] := 1 - Exp[-(x/dm)^n]

(* Log of likelihood *)
logL = Sum[c[[i]] Log[F[x[[i + 1]]] - F[x[[i]]]], {i, Length[c]}];

$$ \log {L} = \sum_ {i = 1}^k c_i \log {(F (x_ {i + 1}) - F (x_i))}$$

(* Find maximum likelihood estimates *)
sol = FindMaximum[logL, {{n, 4}, {dm, 200}}, WorkingPrecision -> 20]
(* {-5891.8004314673819211, {n -> 4.3748213953340505200, dm -> 191.19015070319332793}} *)

(* Show results *)
Show[ContourPlot[logL, {n, 3, 6}, {dm, 150, 250}, PlotRange -> All,
  Contours -> -{6, 7, 8, 9, 10, 11, 12} 1000, 
  PlotLabel -> Style["Log likelihood surface", Black, 24],
  Frame -> True, 
  FrameLabel -> (Style[#, Bold, Italic, 18] &) /@ {"n", "dm"}],
 ListPlot[{{n, dm}} /. sol[[2]], PlotStyle -> Red]]

(* Standard errors *)
cov = -Inverse[(D[logL, {{n, dm}, 2}]) /. sol[[2]]];
sen = cov[[1, 1]]^0.5
(* 0.05702 *)
sedm = cov[[2, 2]]^0.5
(* 0.786466 *)

To repeat: this only works if one can reconstruct the counts as the total count is the sample size rather than the number of bins (which is completely artificial).

Given that the dataset name (sizedistcum) has "dist" and "cum" in it and it appears that you have equal intervals with respect to the log of x, I suspect you have a sample cumulative distribution function formed from a binning of a random sample of size 3,705. The function that you're trying to fit is a cumulative distribution function for a Weibull distribution. If so, performing a regression is inappropriate.

The counts for the histogram seem to be

{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 8, 97, 94, 80, 103, 614, 2089, 488, 108, 24, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}

If so, I can show you how to estimate the parameters of the Weibull distribution from a histogram if my speculation about your data is correct. But first I want to determine if my speculation is correct.

Jim