How many squares in base $9$ consist of only ones How many perfect squares exist of the following form: $(1111111....11)_9$?
Let the required number equal be $(111111...1\space (n \space ones))_9$. This can written as:
$9^{n-1} + 9^{n-2} + \dots + 9^2 + 9 + 1 = \frac{9^{n} - 1}{8} = k^2$ for some k. 
Since $\frac14$ is a perfect square, $\frac{9^n - 1}{2}$ must be a perfect square as well.
So, the problem boils down to finding all $n$ such that $\frac{9^n - 1}{2}$ is a perfect square.  
$\frac{(3^n - 1)(3^n + 1)}{2}$
One such solution seems to be $n = 1$ which yields $4$. This was the only small solution I was able to find by hand. There are, however, others, which I found out when I ran a small computer program of $n$s upto $100$: $33,
34,
35,
36,
37,
38,
39,
40,
41,
42,
43,
44,
45,
46,
47,
48,
49,
50,
51,
52,
53,
54$ amongst others. 
Is there any way to identify these by hand?
EDIT: According to the comments, the numerical findings are incorrect. Indeed, upon checking the program, I found that this was the case. 
Now, it only remains to be proven that $n=1$ is the only solution. How do I do that? I would prefer an elementary proof. 

The following is to show there are not solutions with $n>1$ of the equation.
The equation $(9^n-1)/8=k^2$ may, after doubling the sides, be factored as
$$\frac{3^n-1}{2}\cdot \frac{3^n+1}{2}=2k^2.$$
The two factors on the left here are adjacent integers and so are coprime. This implies, from the right side of the equation, that there are $u,v$ such that one factor is $2u^2$ and the other is $v^2.$ One gets something by considering the factor which is of the form $2u^2.$
Consider the possibility that $(3^n-1)/2=2u^2.$ This implies that $3^n-1=2\cdot 2u^2=(2u)^2.$ If $n>1$ this is impossible, since by Catalan's "conjecture", (now a theorem; see below) the only nontrivial perfect powers differing by $1$ are $8$ and $9$.
On the other hand it may be that $(3^n+1)/2=2u^2$, which implies $3^n+1=(2u)^2$, and again we get if $n>1$ two perfect powers other than $8,9$ which differ by $1$ against Catalan's conjecture.
Link for the Catalan conjecture/theorem:
http://en.wikipedia.org/wiki/Catalan%27s_conjecture
ADDED: The OP has asked for a simpler proof, not using advanced theorems. I believe the following fits the bill, only based on simple congruence ideas.
For this we need to use the "parity fact"  that $(3^n-1)/2$ is odd or even when $n$ is respectively odd or even. The latter can be shown by noting that from $3^{2k}-1=8r$ follows $(3^{2k}-1)/2=4r,$ and multiplying by $3$ and rearranging, $3^{2k+1}-1=3\cdot8r+2,$ so that $(3^{2k+1}-1)/2=3 \cdot 4r+1.$
Recall from the above that we have the two adjacent integers $(3^n-1)/2,\ (3^n+1)/2$ and know that one of them is of the form $2k^2$. If $n$ is odd this is $(3^n+1)/2$ from the parity fact, while if $n$ is even it is the factor $(3^n-1)/2$ which is of form $2k^2$.
Consider the case with $n$ odd and $3^n+1=(2k)^2.$ From this, we have
$$3^n=(2k-1)(2k+1),$$
where the factors on the right are coprime since they are odd and differ by $2$. By unique factorization, it follows that the smaller factor must be $1$ (otherwise we factor $3^n$ into nontrivial coprime factors). This leads to $k=1$ and then to $n=1$.
Consider on the other hand the case with $n$ even and $3^n-1=(2k)^2.$ Since $n \ge 1$ and $n$ is even, we have $n=2m$ with $m\ge 1$, and also
$$(3^m)^2-1=(2k)^2,$$
which is two squares differing by $1$, but the only such squares are $0,1$, whereas here one of the squares is at least $3^2$ since $m \ge 1.$
