Firstly observe that $m$ is odd because if $p|a^{2}+1$ we have $p\equiv 3 (mod 4)$ which is contradiction.Now we have $a\equiv - 1 (mod 15)$,so $2|n$ so $(a^{g}+1)\cdot(a^{g}+1)|15^{k}$ so $a^{g}=2$ so $a=2$ and now we have $2^{m}+1=5\cdot 15^{k-1}$ and we have $2^{m}\equiv 4 (mod 5)$ so $2|m$ which is contradiction.

Garfield wrote: Firstly observe that $m$ is odd because if $p|a^{2}+1$ we have $p\equiv 3 (mod 4)$ which is contradiction.Now we have $a\equiv - 1 (mod 15)$,so $2|n$ so $(a^{g}+1)\cdot(a^{g}+1)|15^{k}$ so $a^{g}=2$ so $a=2$ and now we have $2^{m}+1=5\cdot 15^{k-1}$ and we have $2^{m}\equiv 4 (mod 5)$ so $2|m$ which is contradiction. why is $p\equiv 3 (mod 4)$ a contradiction? actually the equation has solutions, i will post them if nobody posts a solution

ochoa wrote: Garfield wrote: Firstly observe that $m$ is odd because if $p|a^{2}+1$ we have $p\equiv 3 (mod 4)$ which is contradiction.Now we have $a\equiv - 1 (mod 15)$,so $2|n$ so $(a^{g}+1)\cdot(a^{g}+1)|15^{k}$ so $a^{g}=2$ so $a=2$ and now we have $2^{m}+1=5\cdot 15^{k-1}$ and we have $2^{m}\equiv 4 (mod 5)$ so $2|m$ which is contradiction. why is $p\equiv 3 (mod 4)$ a contradiction? actually the equation has solutions, i will post them if nobody posts a solution Yeah I missed some cases,but it's just case bash anyway: 1.st case Let $3$ doesn't divide $a^{m}+1$ so $a^{m}+1=5^{s}$ which has no solution by $Catalan's conjecture$ unless $m=1$ so $a+1=5^{s}$ now if $a^{n}-1=5^{k-s}\cdot 3^{s}$ if $k$ isn't equal $s$ we get case I solved because then $2|n$,so $a^{n}-1=3^{s}<5^{s}<a+1$ ,no solution unless $a=4$ and that's same solution we will get later.Second option is that $a^m +1=5$ so if $a=4$ so $n=1$ (just different od squares) so $(a,m,n,k)=(4,1,1,1)$. 2nd case If $5$ doesn't divide $a^{m}+1$ so $a^{m}+1=3^{s}$ which doesn't have solutions unless $s=1$,$a=2$ now $2^{n}-1=5^{s}\cdot 3^{s-1}$ so $2|n$ so no solutions .It may have solutions in case $a+1=3^{s}$ again we get $s=k$ so $a+1=3^{s}$ and $a-1=5^s$ so contradiction 3rd case $15|a^{m}+1$ and I solved this case as a original problem. So $(a,m,n,k)=(4,1,1,1)$

PS

ochoa wrote: Garfield wrote: Firstly observe that $m$ is odd because if $p|a^{2}+1$ we have $p\equiv 3 (mod 4)$ which is contradiction.Now we have $a\equiv - 1 (mod 15)$,so $2|n$ so $(a^{g}+1)\cdot(a^{g}+1)|15^{k}$ so $a^{g}=2$ so $a=2$ and now we have $2^{m}+1=5\cdot 15^{k-1}$ and we have $2^{m}\equiv 4 (mod 5)$ so $2|m$ which is contradiction. why is $p\equiv 3 (mod 4)$ a contradiction? actually the equation has solutions, i will post them if nobody posts a solution Fermat's theorem on sum of squares.. It states that a prime can only be represented as a sum of squares iff $ p \equiv 1 (mod 4)$ The mistake was after that contradiction i think

there is one more solution $(2,2,2,1)$.

ochoa wrote: there is one more solution $(2,2,2,1)$. Yeah in first case I forgotten subcase when $Catalan's conjecture$ doesn't work $a^m+1=5$ i puted only $a=4$,$m=1$ as a trivial solution ,but i missed $a=2$,$m=2$ which is a solution too $2^n-1=3$ so $n=2$ so $(a,m,n,k)=(2,2,2,1)$.

Well

And actually I do not think you are allowed to use $Catalan's\ conjecture$ in competitions

I thought that anything has name and has been proved can be used on competition.The best example is USAMO 2016 2many students used Hook length formula and get 7 and it trivializes problem.$\text{Catalan's conjecture}$ didnt completly trivialized the problem so I guess it's ok.

Garfield wrote: I thought that anything has name and has been proved can be used on competition.The best example is USAMO 2016 2many students used Hook length formula and get 7 and it trivializes problem.$\text{Catalan's conjecture}$ didnt completly trivialized the problem so I guess it's ok. no, there was a discussion about that and 90% of the students that used the hook lenght lemma got 0 or 1 point... evan chen made a post about it, i''ll look for it

Solved in Portuguese here.