Alternate

Confirmed that $101$ is the only one, using $Mathematica$ , up to $1010...101$ => $3000$ $digits$

here is the program if you want to search further

Monitor[For[n=1,n<10000,n++,If[PrimeQ@FromDigits@Join[{1},Flatten@Table[{0,1},n]],Print@n]],n]

Define the $n^{th}$ term of the sequence to be $a_n$ , so that $a_1=1$ , $a_2=101$ , $a_3=10101$ , etc, and the decimal expansion of $a_n$ contains $n$ $1$ s. We want to show that $a_2=101$ is the only prime in the sequence. The simplest thing to do is to try factorising, and look for patterns.

One idea is to note that when $n$ is a multiple of $3$ , by the usual divisibility test, so is $a_n$ . Similarly, if $n$ is a mulitple of $11$ , so is $a_n$ . But this approach doesn't get too far - there are still lots of gaps.

Playing around with the factorisations, we can find two patterns. For even $n$ :

$a_2=101=101 \times 1$

$a_4=1010101=101 \times 10001$

$a_6=101010101=101 \times 100010001$

and it's easy to see that this pattern continues.

For odd $n$ :

$a_1=1=1 \times 1$

$a_3=10101=91 \times 111$

$a_5=101010101=9091 \times 11111$

$a_7=1010101010101=909091 \times 1111111$

This pattern is slightly less obvious, but still fairly easy to prove.

If we define $r_1=1$ , $r_2=11$ , $r_3=111$ , etc, we conjecture from the pattern that $r_n$ divides $a_n$ whenever $n$ is odd.

Note that each of the $a_n$ and $r_n$ can be thought of as sums of geometric progressions. This gives the formulas $a_n=\frac{1}{99} \left(100^n-1 \right)$ and $r_n=\frac{1}{9} \left(10^n-1 \right)$ .

So we want to show that $\frac{a_n}{r_n}$ is an integer for odd $n$ .

We have $\begin{aligned} \frac{a_n}{r_n} &= \frac{100^n-1}{11 \left(10^n-1 \right)}\\ &=\frac{10^{2n}-1}{11 \left(10^n-1 \right)}\\ &= \frac{\left(10^{n}-1\right)\left(10^{n}+1\right)}{11 \left(10^n-1 \right)}\\ &= \frac{10^{n}+1}{11} \end{aligned}$

and since $n$ is odd, we can see this last form is indeed an integer (using the fact that $x+1$ divides $x^n+1$ when $n$ is odd).

So we've proved that the only prime in the sequence is $101$ .