Determine all positive integer solution sets $(k, n, l, m)$ to the equation $(1 + n^k)^l=1+n^m$ where $l>1$ Determine all positive integer solution sets $(k, n, l, m)$ to the equation
$(1 + n^k)^l=1+n^m$ where $l>1$
I worked out that (1,2,2,3) is a solution and is the only solution where $n=l=2$ because if $n=l=2$ then we have the equation 
$(1+2^k)^2=1+2^m$ 
$2^{2k}+2(2^k)+1=2^m+1$
which simplifies to
$2^{2k}+2^{k+1}=2^m$
which means $k+1=2k$ because $k$ and $m$  are integers
which gives $k=1$
therefore
$m=3k=3$
and I think that is the only solution but Im not sure how to prove this
solutions, suggestions and hints would all be appreciated
from the 2018 South African Senior Team Competition http://www.samf.ac.za/content/files/QuestionPapers/2018_Senior_Team_COMBO.pdf
 A: You're asking for positive integer solutions, with $l > 1$, to
$$(1 + n^k)^l=1+n^m \tag{1}\label{eq1}$$
First, if $n = 1$, then \eqref{eq1} becomes
$$\begin{equation}\begin{aligned}
(1 + 1)^l & = 1 + 1^m \\
2^l & = 2
\end{aligned}\end{equation}\tag{2}\label{eq2}$$
This gives $l = 1$, but as this isn't allowed, it means $n \gt 1$.
Next, expanding the LHS of \eqref{eq1} using the Binomial theorem, you'll get at least $3$ terms. In particular, you have
$$\begin{equation}\begin{aligned}
1 + ln^k + \frac{l(l - 1)}{2}n^{2k} + \sum_{i=3}^{l}\binom{l}{i}n^{ik} & = 1 + n^m \\
ln^k + \frac{l(l - 1)}{2}n^{2k} + \sum_{i=3}^{l}\binom{l}{i}n^{ik} & = n^m \\
l + \frac{l(l - 1)}{2}n^{k} + \sum_{i=3}^{l}\binom{l}{i}n^{(i-1)k} & = n^{m-k} \\
\end{aligned}\end{equation}\tag{3}\label{eq3}$$
On the LHS, from the second term on, each term has a factor of $n^k$. Since each term is positive, their sum is $\gt n^k$, so $n^{m-k} \gt n^k \implies m - k \gt k$. Thus, this means that $n^k \mid l$. Let $i$ be the highest power of $n$ which divides $l$. Then let the highest power of $n$ which divides $\frac{l(l - 1)}{2}n^{k}$ be $r$, with it being $(i - 1) + k$ if $2 \mid n$, else it's $i + k$. If $i$ and $r$ don't match, i.e., $i \lt r$, as $i$ is less than the highest power of $n$ which divides any next term(s) (since each next term has, cumulatively, the binomial coefficient with at most $1$ less factor of $n$ on average, but with $n^{(i-1)k}$ having $k$ more factors of $n$), you can divide both sides of \eqref{eq3} by $n^i$ and have a remaining factor among the sum of the first $2$ terms which is not a factor of $n$, but with the additional terms having at least one factor of $n$. However, this isn't allowed as the RHS is just a power of $n$. This shows $r = i \implies i = i - 1 + k \implies k = 1$. Also, it requires $2 \mid n$ and, since you can show there are no additional factors of $n$ using basically the same argument as above, it's actually $n = 2$. As such, $l$ is even, say $l = 2s$. \eqref{eq1} then becomes
$$\begin{equation}\begin{aligned}
(1 + 2)^l & = 1 + 2^m \\
3^{2s} - 1 & = 2^m \\
(3^{s} - 1)(3^{s} + 1) & = 2^m
\end{aligned}\end{equation}\tag{4}\label{eq4}$$
Both $3^{s} - 1$ and $3^{s} + 1$ are powers of $2$, but only $2$ apart. This only occurs if $3^s - 1 = 2$ and $3^s + 1 = 4$, i.e., $s = 1$ giving $l = 2$ and $8 = 2^m \implies m = 3$.
In summary, this shows the only solution is the one you found of $(1,2,2,3)$.
