Why a $20$ digit number starting with eleven $1$'s cannot be a perfect square?

That's because\begin{align*}3\,333\,333\,333^2&<11\,111\,111\,111\,000\,000\,000\\&<11\,111\,111\,111\,999\,999\,999\\&<3\,333\,333\,334^2.\end{align*}

Adding onto the answer by José Carlos Santos, here's a way to compute the approximate square root by hand efficiently:

$$\sqrt{11111111111000000000} < \sqrt{11111111111111111111 + \frac{1}{9}} < \sqrt{11111111111999999999}$$ $$\sqrt{11111111111111111111 + \frac{1}{9}} = \sqrt{\frac{10^{20}}{9}} = \frac{10^{10}}{3} = 3333333333 + \frac{1}{3}$$

Then, how to check that the neighboring squares fall outside the desired range, by hand:

\begin{align*} \left\lfloor\frac{10^{10}}{3}\right\rfloor^2 &= \left(\frac{10^{10}}{3} - \frac{1}{3}\right)^2 \\ &= \left(\frac{10^{10}}{3}\right)^2 - 2\left(\frac{10^{10}}{3}\right)\left(\frac{1}{3}\right) + \left(\frac{1}{3}\right)^2 \\ &= \frac{10^{20}}{9} - \frac{2\times10^{10} - 1}{9} \\ &< \frac{10^{20}}{9} - \frac{10^9}{9} \\ &= 11111111111000000000 \end{align*}

\begin{align*} \left\lceil\frac{10^{10}}{3}\right\rceil^2 &= \left(\frac{10^{10}}{3} + \frac{2}{3}\right)^2 \\ &= \left(\frac{10^{10}}{3}\right)^2 + 2\left(\frac{10^{10}}{3}\right)\left(\frac{2}{3}\right) + \left(\frac{2}{3}\right)^2 \\ &> \frac{10^{20}}{9} + \frac{4\times10^{10}}{9} \\ &> \frac{10^{20}}{9} + \frac{8\times10^9}{9} \\ &= \frac{10^{20}}{9} - \frac{10^9}{9} + 10^{9}\\ &= 11111111111000000000 + 10^{9} \\ &> 11111111111999999999 \end{align*}

Let's show more generally that a $2n$-digit square cannot begin with $n+1$ $1$'s.

Note that the smallest $2n$-digit number beginning with $n+1$ $1$'s is

$$11{\ldots}110{\ldots}0={10^{2n}-10^{n-1}\over9}$$

and the largest is

$$11{\ldots}119{\ldots}9={10^{2n}-10^{n-1}\over9}+10^{n-1}-1={10^{2n}+8\cdot10^{n-1}-9\over9}$$

Since $10^{n-1}\lt2\cdot10^n-1$ while $8\cdot10^{n-1}-9\lt2\cdot10^n+1$, we obtain

$$\left(10^n-1\over3 \right)^2={10^{2n}-2\cdot10^n+1\over9}\lt{10^{2n}-10^{n-1}\over9}=11{\ldots}110{\ldots}0$$

while

$$11{\ldots}119{\ldots}9={10^{2n}+8\cdot10^{n-1}-9\over9}\lt{10^{2n}+2\cdot10^n+1\over9}=\left(10^n+1\over3\right)^2$$

Thus if $11{\ldots}110{\ldots}0\le N\le11{\ldots}119{\ldots}9$, then

$${10^n-1\over3}\lt\sqrt N\lt{10^n+1\over3}\lt{10^n+2\over3}$$

But $(10^n-1)/3$ and $(10^n+2)/3$ are consecutive integers (their difference is $1$). So $\sqrt N$ cannot be an integer.

Remark 1: This answer owes much of its logic to a now-deleted answer by achille hui.

Remark 2: This answer was motivated by JollyJoker's first comment beneath the OP. JJ's follow-up comment there makes it clear that squares with $2n+1$ digits can begin with more than $n+1$ $1$'s.