Note that $n$ must always clearly be even. We now split the problem into two cases. Case 1. Consider $k$ to be even. Then $3^k+5^k\equiv 2\pmod{4}$. But since $m\geq 2$ and $n\equiv 0\pmod{2}$, we have that $n^m\equiv 0\pmod{4}$, a clear contradiction. Case 2. If $k$ is odd, it is easy to see that $$v_2(n^m)=v_2(3^k+5^k)=3$$But it's clear that $m\mid v_2(n^m)$ and since $m\geq 2$, this implies that $m=3$. If $k=1$ we obtain the solution $(k, n, m)=(1, 2, 3)$. If $k\geq 2$ then $3^k\equiv 0\pmod{9}$ and furthermore, recalling that $k$ is odd, $5^k\equiv 5, 8, 2\pmod{9}$, and therefore we have $n^3=3^k+5^k\equiv 5, 8, 2\pmod{9}$. Since it's clear that it's only possible that $n^3\equiv 0, 1, 8\pmod{9}$, it follows that we must have $5^k\equiv 8\pmod{9}$ which can only happen if $3\mid k$, so we can write $k=3t$ for some $t\in\mathbb{N^{*}}$. Denoting $3^t=a$ and $5^t=b$, our equation rewrites as $$a^3+b^3=n^3$$which can be proven to be imposibble either by directly citing Fermat's Last Theorem, or by looking modulo $7$ (without the notations). We conclude that the only valid solution is $(k, n, m)=(1, 2, 3)$.

Note that $n$ is even. Now if $k$ is odd then, we get $v_2(3^k+5^k)=1\implies m=1.$ Not possible, so $k$ is odd. Now using LTE we get $v_2(3^k+5^k)=v_2(3+5)=3\implies m=3.$ For $k=1,$ we get $(k, n, m)=(1, 2, 3).$ Assume $k>1.$ Taking $\mod 9,$ we get $5^k\equiv 1,-1 \mod 9 \implies 3|k. $ But now by $\mod 7$ we get a contradiction.