What is the relative size of the (M+2) peak?
Let’s assume your compound is $\ce{C_nH_xO_y}$. Thankfully, both hydrogen and oxygen are elements that only have one significant naturally occuring isotope. Therefore, we can treat the entire contribution of $\ce{H_xO_y}$ as a constant $c$. All we need to answer is how large the $M+1$ and $M+2$ peaks are given the number of carbons, $n$. (This treatment is not entirely correct as highlighted by orthocresol, but it is close enough for my purposes.)

Assuming you do not have any isotope enrichments, each carbon will independently have a chance of either being $\ce{^12C}$ or $\ce{^13C}$ (again, we will ignore all other isotopes such as $\ce{^14C}$). The independence is the big key word here. We can use general principles of stochastics to calculate the result.

The probability of all carbon atoms of one molecule of your compound are $\ce{^12C}$ is:
$$P(\ce{^12C_nH_xO_y}) = 0.989^n$$
The $M+1$ peak is represented by one single atom being $\ce{^13C}$. Again, the principles of stochastics apply:
$$P(\ce{^12C_{n-1}^13C1H_xO_y}) = \left ( n\atop 1\right) \times 0.989^{n-1} \times 0.011$$
And finally for the $M+2$ peak:
$$P(\ce{^12C_{n-2}^13C2H_xO_y}) = \left ( n \atop 2\right) \times 0.989^{n-2} \times 0.011^2$$

Using this, we can plot what the probability of each peak is – and use that as their relative heights.



We see that from $n=90$ the $M+1$ peak is actually larger than the $M$ peak. From $n=128$, even the $M+2$ peak will be larger than the $M$ peak. And from $n=181$ the $M+2$ peak becomes the largest of these three.

Of course, further peaks will also start appearing, meaning that the mass spectra of very large molecules will become difficult to analyse.

Now the numbers from which $M$ is no longer te principal peak are quite alrge. Unless you are synthesising maitotoxin, you will probably not need to resort to any analysis that does not only involve $M+1$.