Why are there specifically 10, 11, or 26 dimensions in string theory? I know that current string theories state that there are 10, 11, or 26 spacetime dimensions in superstring theory, M-theory, and bosonic string theory, respectively. But when I looked up why those numbers were chosen and not, say, 35, I could not find an answer that suited me.
As I am not an expert in string theory, can you please explain to me why (if possible in easier terms) those specific answers were chosen in regard to the number of dimensions?
 A: In bosonic string theory, after canonical quantisation, we can construct particle states from the creation operators $\tilde{\alpha}^i_{-1}$  and $\alpha^j_{-1}$, as $\tilde{\alpha}^i_{-1} \alpha^j_{-1} |0; p\rangle$, which turn out to each have a mass of,
$$M^2 = \frac{4}{\alpha'}\left( 1- \frac{d-2}{24}\right)$$
where $d$ is the space-time dimension and $\alpha'$ is the Regge slope. Why is this a problem? In short, Wigner's classification of representations of the Poincaré group says any massive particle must form a representation of $SO(d-1)$ in $\mathbb{R}^{1,d-1}$. Without going into the entire canonical quantisation, the particles described have $(d-2)^2$ states. However, if these particles happen to be massless, they are permitted and have fewer internal states than massive particles, just like how a photon in four dimensions has two degrees of freedom - its two polarisations.
Thus, for them to be massless and preserve Lorentz symmetry, we must require $M^2=0$, so $d=26.$

There is another way to view the reason for $d=26$ in bosonic string theory. When examining the CFT given by the Polyakov action, one of the conditions of a CFT one can derive is,
$$T^\alpha_\alpha = 0$$
that is, the trace of the stress-energy tensor for the theory vanishes. However, when quantising the theory, it turns out that,
$$\langle T^\alpha_\alpha\rangle  = -\frac{c}{12}R$$
where $c$ is the central charge and $R$ is the Ricci scalar of the curved background of our theory. This is known as an anomaly, and it arises in 'simpler' theories, like the chiral anomaly of quantum electrodynamics.
Now, when quantising the CFT using the path integral, ghost particles necessarily arise, just like they do when quantising non-Abelian gauge field theories. It turns this ghost system has $c = -26$. So, for the whole theory to have $c = 0$ and thus keep Weyl symmetry, we need to add $26$ scalar fields which each contribute $c = 1$. 

Although you've asked for a non-technical description, the reality is these conditions stem from specific mathematical details of the theory, and more specifically from the symmetries they must possess.
