First, note that the function $\iota(m) = m$ is such a function, as
$$(\iota(m))^2 + 2m\iota(n) + \iota(n^2) = (m+n)^2.$$

We show that $\iota$ is the only such function.
Let $f$ be any such function. Substituting $m = n = 1$, we see that $f(1)^2 + 3f(1)$ must be a square. As $(f(1)+1)^2 \le f(1)^2 + 3f(1) < (f(1)+2)^2$, the inequality on the left must be an equality, so $f(1) = 1$.

Now let $k$ be any positive integer for which $p = 2k + 1$ is an odd prime. By substituting $m = k = \frac{1}{2}(p-1)$ and $n = 1$, we see that $f(k)^2 + p$ must be a square, say $a^2$ for a positive integer $a$. Then $p = a^2 - f(k)^2 = (a-f(k))(a+f(k))$, so by unique factorisation and $a-f(k) < a+f(k)$ it follows that $a-f(k) = 1$ and $a+f(k) = p = 2k+1$. Taking the difference of these two equalities yields $f(k) = k$.
Now let $x$ be any positive integer, and choose a positive integer $k$ for which $p = 2k + 1$ is an odd prime (so that $f(k) = k$) and that is large enough so that $1-2k-2f(x) < f(x^2)-f(x)^2 < 1+2k+2f(x)$. Substituting $m = k$ and $n = x$ then gives that $k^2+2kf(x)+f(x^2) = (k+f(x))^2+f(x^2)-f(x)^2$ must be a square. By our chosen bound on $k$, we have
$$(k + f(x) - 1)^2 < (k + f(x))^2 + f(x^2) - f(x)^2 < (k + f(x) + 1)^2.$$
The expression in the middle is a square and therefore must equal $(k+f(x))^2$, therefore $f(x^2) = f(x)^2$ for all positive integers $x$.

Finally, let $x$ be any positive integer, and choose a positive integer $k$ for which $p = 2k + 1$ is again an odd prime and that is large enough so that $1-2x-2k < (f(x))^2 - x^2 < 1+2x+2k$. Substituting $m = x$ and $n = k$ then gives that $(f(x))^2 + 2xk + k^2 = (x+k)^2 + (f(x))^2 - x^2$ must be a square. By our chosen bound on $k$, we have
$$(x + k - 1)^2 < (x + k)^2 + (f(x))^2 - x^2 < (x + k + 1)^2.$$
The expression in the middle is a square and therefore must equal $(x+k)^2$, therefore $(f(x))^2 = x^2$ for all positive integers $x$.
Thus $f(x) = x$ for all positive integers $x$, so $f$ must be equal to $\iota$. Therefore $\iota$ is the only function satisfying the required property. ☐