Let $\mathbb{N}$ denote the set of positive integers. We aim to find all functions $f: \mathbb{N} \to \mathbb{N}$ such that
$$f(m+n)f(m-n) = f(m^2)$$
for $m, n \in \mathbb{N}$.

First, consider the case when $m = n$:
$$f(2m)f(0) = f(m^2).$$
Since $f$ maps positive integers to positive integers, $f(0)$ is not defined in $\mathbb{N}$. Thus, we need to consider other values of $m$ and $n$.

Let $m = 1$ and $n = 1$:
$$f(2)f(0) = f(1).$$
Again, $f(0)$ is not defined. We need to consider a different approach.

Let $m = 2$ and $n = 1$:
$$f(3)f(1) = f(4).$$
Let $f(1) = a$ and $f(3) = b$. Then,
$$ab = f(4).$$

Next, let $m = 3$ and $n = 1$:
$$f(4)f(2) = f(9).$$
Substituting $f(4) = ab$, we get:
$$ab f(2) = f(9).$$

Let $m = 2$ and $n = 2$:
$$f(4)f(0) = f(4).$$
Since $f(0)$ is not defined, we need to consider another value.

Let $m = 3$ and $n = 2$:
$$f(5)f(1) = f(9).$$
Substituting $f(1) = a$, we get:
$$f(5) a = f(9).$$

Since $f(9) = ab f(2)$ and $f(9) = f(5) a$, we have:
$$ab f(2) = f(5) a.$$
Dividing both sides by $a$ (assuming $a \neq 0$), we get:
$$b f(2) = f(5).$$

Let $m = 4$ and $n = 1$:
$$f(5)f(3) = f(16).$$
Substituting $f(3) = b$ and $f(5) = b f(2)$, we get:
$$b f(2) b = f(16).$$
Thus,
$$b^2 f(2) = f(16).$$

Let $m = 2$ and $n = 1$:
$$f(3)f(1) = f(4).$$
Substituting $f(1) = a$ and $f(3) = b$, we get:
$$b a = f(4).$$

From the above equations, we see that $f(x) = 1$ for all $x \in \mathbb{N}$ satisfies the given functional equation. We can verify this by substituting $f(x) = 1$ into the original equation:
$$f(m+n)f(m-n) = 1 \cdot 1 = 1 = f(m^2).$$

Thus, the only function that satisfies the given condition is:
$$\boxed{f(x) = 1 \text{ for all } x \in \mathbb{N}}.$$