$m^4+4^kp^4|m^2(m^4-4^kp^4)\implies m^4+4^kp^4|m^2(2\cdot4^kp^4)$ $\gcd(m^4+4^kp^4,2)=1 \implies m^4+4^kp^4|m^2p^4$ FTSOC, assume that $\gcd(m,p)=1$. Since $\gcd(m^4+4^kp^4,m)=\gcd(4^kp^4,m)=\gcd(p^4,m)=1$ and $\gcd(m^4+4^kp^4,p)=\gcd(m^4,p)=1$, $\gcd(m^4+4^kp^4,m^2p^4)=1$. So, $m^2p^4=\pm 1$ which is absurd since $p$ is prime. Thus $\gcd(m,p)=p$. Let $m=pd$ with $d\in\mathbb{Z}^+$. Since $m$ is odd, $d$ must also be odd. Let $k=2q-1$ with $q\in\mathbb{Z}^+$. $m^4+4^kp^4|m^2p^4\implies p^4d^4+2^{4q-2}p^4|p^6d^2\implies d^4+2^{4q-2}|p^2d^2$ Since $\gcd(d^4+2^{4q-2}, d^2)=\gcd(2^{4q-2}, d^2)=1$, $d^4+2^{4q-2}|p^2$. $\boxed{d^4+2^{4q-2}=1,p,p^2}$ Case 1: $d^4+2^{4q-2}=1$ $2^{4q-2}\geq 1\implies d^4\leq 1-1=0$ which is absurd. Case 2: $d^4+2^{4q-2}=p$ $(d^2-2^qd+2^{2q-1})(d^2+2^qd+2^{2q-1})=(\pm 1)(\pm p)=(\pm p)(\pm 1)$ Since $d^2+2^qd+2^{2q-1}>d^2-2^qd+2^{2q-1}, d^2+2^qd+2^{2q-1}+d^2-2^qd+2^{2q-1}>0$, we have $d^2-2^qd+2^{2q-1}=1, d^2+2^qd+2^{2q-1}=p$. $2d^2-2^{q+1}d+2^{2q}=2\implies (d-2^q)^2=2-d^2\implies 2-d^2\geq 0\implies d=1$ $(d-2^q)^2=2-d^2\implies (1-2^q)^2=2-1^2\implies 1-2^q=\pm 1\implies 2^q=0$ or $2$ $q=1\implies k=2\cdot 1-1=1$ $d^4+2^{4q-2}=p\implies 5=1^4+2^{4\cdot 1-2}=p$ $m=pd=5\cdot 1=5\implies (p,m,k)=(5,5,1)$ Case 3: $d^4+2^{4q-2}=p^2$ $(d^2-2^qd+2^{2q-1})(d^2+2^qd+2^{2q-1})=(\pm 1)(\pm p^2)=(\pm p^2)(\pm 1)=(\pm p)(\pm p)$ Since $d^2+2^qd+2^{2q-1}>d^2-2^qd+2^{2q-1}, d^2+2^qd+2^{2q-1}+d^2-2^qd+2^{2q-1}>0$, we have $d^2-2^qd+2^{2q-1}=1, d^2+2^qd+2^{2q-1}=p^2$. $2d^2-2^{q+1}d+2^{2q}=2\implies (d-2^q)^2=2-d^2\implies 2-d^2\geq 0\implies d=1$ $(d-2^q)^2=2-d^2\implies (1-2^q)^2=2-1^2\implies 1-2^q=\pm 1\implies 2^q=0$ or $2\implies q=1$ $d^4+2^{4q-2}=p^2\implies 5=1^4+2^{4\cdot 1-2}=p^2$ which is absurd. So, $(p,m,k)=(5,5,1)$.

Standard. See that the condition is equivalent to solve \[m^4+4^kp^4 \mid 2m^2 \gcd(m^4,4^kp^4)\]Claim: We have $n^4+4^k$ as a power of $p$ where $m=np$. Call it $p^t$ with $t \geq 1$. Proof: As LHS $>1$ we can say if a prime $q$ divides it (has to be odd) then $q \mid 2m^6 \iff q \mid m \implies q \mid p$ and hence LHS is a power of $p$. Obviously $p \mid m$ and the result follows. $\square$ Claim: We have $t=1$ or $2$. Proof: See that $p^{t+4} \mid 2n^6p^6$ and if $p \mid n$ then $p \mid 4^k$, contradiction. Hence $t+4 \leq 6 \iff t \leq 2$. $\square$ Case A: $t=1$ Write it as \[n^4+4 \cdot 16^{\frac{k-1}2}=p \iff (n^2+2n \cdot 16^{\frac{k-1}2}+2 \cdot 16^{k-1})(n^2-2n \cdot 16^{\frac{k-1}2}+2 \cdot 16^{k-1})=p\]This means that if we let $\ell=16^{\frac{k-1}2}$ then \[(n-\ell)^2+\ell^2=1 \iff n=\ell=1\]This means that $(p,m,k)=(5,5,1)$ which obviously works. Case B: $t=2$ Write it as \[4^k=(p-n^2)(p+n^2)\]Both must be even and hence both are powers of $2$ and if both are divisible by $4$ then their sum which is $2p$ is divisible by $4$, contradiction. So $p=n^2+2$ and so $n^2+1=4^{k-1}$ and if $k \geq 2$ then $4 \mid n^2+1$, contradiction and so $k=1$ but this means $n=0$. And so all in all, the only solution is $\boxed{(p,m,k)=(5,5,1)}$.