Solution:

Setting $p = m - n$ and $q = m + n$ gives
$$m n = \frac{1}{4}\left(q^{2} - p^{2}\right) \quad \text{and} \quad m^{2} + n^{2} = \frac{1}{2}\left(q^{2} + p^{2}\right)$$
Hence the given conditions can be written as
$$8 l^{2} - 2 q^{2} + 2 p^{2} = q^{2} + p^{2} + 5 q l$$
i.e.
$$p^{2} = (3 q + 8 l)(q - l)$$
Since $p$ is a prime number and $3 q + 8 l > q - l$ we obtain $p^{2} = 3 q + 8 l$ and $1 = q - l$. Hence $11 l + 3 = p^{2}$.