Take mod 15, finding that since $m,k>1$. We have $n^3 \equiv 10 $ $($mod$ 15)$ The only time this happens is when $n \equiv 10 $ $($mod $15)$ i don't know what to do from here

It's the sum of two cubes, so factor. RHS=$(n+5)(n^2-5n+25)$. We see that $(n+5)$ is the product of a power of 3 and a power of 5, and so is $(n+5)^2=n^2+10n+25$. However, $(n+5)^2-15n$ is also the product of a power of 3 and a power of 5. Their gcf is a factor of 15. However, the powers of $(n+5)^2$ are both even, so we have $n^2-5n+25=15$. There seems to be no integer solutions. nvm nvm nvm n can factors of 3 and 5 itself.

Sorry for wrong solution, now second version: $5|n$ so $k\geq 3,n=5a$ $3^m5^{k-3}=a^3+1$ Easy to prove, that $5$ and $9$ not divide $a^2-a+1$ so two cases $a^2-a+1=1,3$ but only $a=2$ is solution. Answer:$(2,3,10)$

@above How do you get $k=3$?

Quite easy problem for tst.

Korgsberg wrote: Quite easy problem for tst. The method of solving $(pq)^a=P(x)$ for some primes $p,q$, positive integer $a$ and a factorizable polynomial $P(x)\in\mathbb{Z}[x]$ where $x$ is an integer is well known. However, problems with innovative ideas are rarely seen in contests.

So let $m=a+b$,$k=c+d$, if we factorize the $RHS$ we get $3^m5^k=(n+5)(n^2-5n+25)$ Because $3$ and $5$ are prime, we have the following: $$n+5=3^a.5^c$$$$n^2-5n+25=3^b.5^d$$Expressing $n$ as $n=3^a.5^c-5$ and putting this into the second equation we get: $$3^{2a}5^{2c}-3^{a+1}5^{c+1}+3.5^2=3^b5^d$$So now let's say one of $a$ or $b$ is $0$, then the equation wouldn't have a solution, so we now know that $a,b > 0$, and we can go forth and divide by $3$. Now the equation looks like this: $$3^{2a-1}5^{2c}-3^a5^{c+1}+5^2=3^{b-1}5^d$$Which obviously implies that $b-1=0 \iff b=1$ Now let's take a look at the factor of $5$, obviously let's say $d$ or $c$ is $0$, then we simply see that we don't have a solution, so divide by $5$ and by the same logic we see that we can divide by $5$ again.We also see that $d=2$.So now we transformed it into the following: $$3^{2a-1}5^{2c-2}-3^a5^{c-1}+1=1$$Which implies that (by factoring): $$3^{a-1}5^{c-1}=1 \iff a-1=0,c-1=0 \iff a=1,c=1$$So now we got that $m=a+b=1+1=2$ and $k=c+d=1+2=3$ and that $n = 15-5=10$. So the only solution is $\boxed{(m,k,n)=(2,3,10)}$

I can confirm what EulersTurban said

some minor cases can be simply deduced from mod3 and mod 5, but some cases come from the Zsigmondy theorem.

Clearly, RHS is divisible by $5$ and so $n$ is divisible by $5$. Let's substitute $n=5a$ and divide both sides by $125$. This yields $3^m5^{k-3} =a^3+1$. It is obvious that if $a^3\equiv2 (mod 3)$ and $a^3\equiv4 (mod 5)$, then $a\equiv2 (mod 3)$ and $a\equiv4 (mod 5)$. So, if 3 or 5 divide $a^3+1$, they should also divide $a+1$. Now we can use the LTE lemma. $m=v_{3}(a^3+1)=v_{3}(a+1)+v_{3}(3)=v_{3}(a+1)+1$ $k-3=v_{5}(a^3+1)=v_{5}(a+1)+v_{5}(3)=v_{5}(a+1)$ Now factoring the expression: $a^3+1=(a+1)(a^2-a+1)$ We know $a+1$ has $m-1$ factors of $3$ (only one less then $a^3+1$) and $k-3$ factors of $5$ (the same as $a^3+1$). So this leaves the other factor, $a^2-a+1$, exactly one factor of 3 and nothing else. Note that it also cannot have other prime factors because then those same primes would divide $3^m5^k$. Thus, $a^2-a+1=3$. $a=2$. This yields $(m,k,n)=(2,3,10)$.

nimueh wrote: Let $m,k,n$ be positive integers. Determine all triples $(m,k,n)$ satisfying the following equation: $3^m5^k=n^3+125$ $3^m5^k \equiv n^3+125\pmod{5}$ $\Rightarrow n\equiv 0 \pmod{5}$ $\Rightarrow n=5t$ $\Rightarrow 3^m5^{k-3}=t^3+1, t\equiv 1 \pmod{2}$ Let us note that $t>1, gcd(t,1)=1$ By Zsigmondy's Theorem: $\exists$ prime $q\neq 2/ q|t^3+1 \Rightarrow q|3^m5^{k-3}$ If $q=3$: $\Rightarrow 3|t^3+1, 3\nmid t+1$ $\Rightarrow t\not\equiv 2 \pmod{3}$ $\Rightarrow 3\nmid t^3+1 (\Rightarrow \Leftarrow)$ $\Rightarrow q=5:$ $\Rightarrow 5|t^3+1, 5\nmid t+1$ $\Rightarrow t\not\equiv 4\pmod{5}$ $\Rightarrow 5\nmid t^3+1 (\Rightarrow \Leftarrow)$ $\Rightarrow$ there is no solution But remember that the theorem has an exception: $\Rightarrow t=2 \Rightarrow n=10,m=2$ and $k=3$ $\Rightarrow (m,k,n)=(2,3,10)$ is a unique solution $_\blacksquare$