Answer: $(m, n, p) = (1, 1, 2)$.

Setting $k = 2024$, $m_1 = m/d$ and $n_1 = n/d$, the given identity becomes $d^{k-2}(dm_1^{k+1} + n_1^k) = pm_1n_1$, where $d$ is the greatest common divisor $(m, n)$ of $n$ and $m$. Put $a = n_1/b$ and $c = d/b$, where $b$ denotes $(d, n_1)$. Then we get $(cb)^{k-2}(cm_1^{k+1} + a^k b^{k-1}) = pm_1a$. Observe that $cm_1^{k+1} + a^k b^{k-1}$ is greater than 1 and is relatively prime to $m_1a$. Then $(cb)^{k-2}$ is divisible by $m_1a$ and
$$\frac{(cb)^{k-2}}{m_1a}(cm_1^{k+1} + a^k b^{k-1}) = p \quad (0.3)$$
which implies $(cb)^{k-2} = m_1a$. Since $(m_1, b) = 1$ and $(a, c) = 1$ we must have $a = b^{k-2}$ and $m_1 = c^{k-2}$. Thus, it follows from (0.3) that $c^{k^2-k+1} + b^{k^2-k+1} = p$, hence $p$ must be divisible by $b+c$ since $k^2-k+1$ is odd. Therefore $p = c+b$ which gives us $c+b = c^{k^2-k+1} + b^{k^2-k+1}$. Hence we conclude that $c = b = 1$ and so $m = n = 1, p = 2$.