Can you count them all?

Notice $11111....=111...08+3=4k+3$ for some integer $k$ .

By division algorithm $\color{limegreen}{\text{Every integer is of form 4k,4k+1,4k+2 or 4k+3,}}$

Now, ${ (4k) }^{ 2 }={ 16k }^{ 2 }=4j\quad \\ \\ \quad { (4k+1) }^{ 2 }={ 16k }^{ 2 }+8k+1=4p+1\quad \\ { (4k+2) }^{ 2 }={ 16k }^{ 2 }+16k+4=4l\quad \\ { (4k+3) }^{ 2 }={ 16k }^{ 2 }+24k+9=4m+1$

For some $j,p,l,m$

From this we see that every square number is of form $4k,4k+1$ .

Now we have to find such a square number whose form is $4k+3$ .

But we know that every square number is of form $4k,4k+1$ but not of $4k+3$ .

Therefore there are $0$ square numbers in sequence.