To address what you don’t understand about the solution given, you should become aware that there is another rule for divisibility by $7$ besides the one you mentioned. This rule is to alternately add and subtract $3$-digit chunks of the number starting with the last $3$ digits and testing whether the result is divisible by $7$. For example, $7003010$ is divisible by $7$ because $10-3+7$ is. This rule works because $7$ divides $1001.$ (By the way, it works for $11$ and $13$ too.) Using this rule, it becomes obvious that any number written as a string of $n$ $1$s, where $n$ is a multiple of $6$, is divisible by $7$.
In the case of $14444$ base $b\ge5$ there can be no squares. We have
$14400_b<14444_b<14641_b$
$(b^2+2b)^2<14444_b<(b^2+2b+1)^2$
and the left and right members are consecutive squares.
With six 4's again there must be no squares:
$1442401_b<1444444_b<1444804_b$
$(b^3+2b^2+1)^2<1444444_b<(b^3+2b^2+2)^2$
We can go on like this. Consider the Laurent series
$\sqrt{1.4444..._b}=\sqrt{1+\dfrac{4b^{-1}}{1-b^{-1}}}=1+\sum\limits_{k=1}^\infty a_kb^{-k}$
$a_1=2,a_2=0,a_3=2,a_4=-
2,...;a_k\in\mathbb{Z}\text{ for all } k$
If $a_m$ is followed by $m$ zeroes in the coefficient sequence then
$P(m)=b^m\left(1+\sum\limits_{k=1}^m a_kb^{-k}\right)$
will be an exact square root of $1444..44_b$ with $n=2m$ fours in all bases; but this can happen only for $m\in\{0,1\}$ because the the known limitations in base ten. For larger $m$, $P(m)$ becomes one of two consecutive whole numbers that strictly bracket $\sqrt{1444...44_b}$. For example, $P(3)=b^3+2b+2$ and we saw above that for $n=2×3=6$ the target square root lies strictly between $P(3)-1$ and $P(3)$. Thereby there are no squares in any base with an even number $\ge4$ of fours.
Best Answer
If $n\ge 4$, then $2^n$ must be a multiple of $16$, therefore so are its last four digits. There are no factors of $5$ so the last four digits cannot be $0000$. Now, which other possibilities for four identical last digits will give a multiple of $16$? Try them out and see.
The comments identify $2^{39}=...888$. If you work out the part above you are left with what must be the answer.