Let A1 = 14, and let An be the number of the form 1444... with n fours following the 1. Find all n such that An is a perfect square.
It is obvious that A2 = 144 is a perfect square and possibly the only one, yet I haven't found a way to prove that.
Possibly the biggest hint is that if $$ An = (13*10^n - 4) / 9\\$$ and if $$ An = k^2 $$
then $$ k = 13i + 1 \\ or \\ k = 13i + 12 $$ for some value of i.
As you suggested, let $A_{n} = k^2$. You've checked for $n = 1$ and, for $n = 2$, you have $144 = 12^2$. Also, note with $n = 3$ that you have a solution of $1444 = 38^2$. Thus, next consider only cases where $n \ge 4$.
You have
$$\begin{equation}\begin{aligned} k^2 & = \frac{13(10^{n}) - 4}{9} \\ 9k^2 & = 13(5^{n})(2^{n}) - 4 \\ (3k)^2 & = 4((13)(5^{n})(2^{n-2}) - 1) \end{aligned}\end{equation}\tag{1}\label{eq1A}$$
This means $k$ is even, let's say $k = 2j$. You can substitute this into \eqref{eq1A} and divide both sides by $4$ to get
$$\begin{equation}\begin{aligned} (3j)^2 & = (13)(5^{n})(2^{n-2}) - 1 \end{aligned}\end{equation}\tag{2}\label{eq2A}$$
For $n \ge 4$, you have $4$ divides $2^{n-2}$. This means that $(3j)^2$ has a remainder of $-1$, which is equivalent to $3$, when divided by $4$. However, all perfect squares have a remainder of only $0$ or $1$ when divided by $4$. This shows there are no solutions for $n \ge 4$, with the only solutions being for $n = 2$ and $n = 3$.
Regarding your hint that $k = 13i + 1$ or $k = 13i + 12$, I don't know of any way to use it to determine the set of valid values of $n$.
