7. Find all functions $f:(0,+\infty) \rightarrow(0,+\infty)$, such that for any positive real numbers $x, y$, we have
$$
\begin{array}{l}
f(f(x)+y) \\
=f(f(x))+2 y f(x)-f(y)+2 y^{2}+1 .
\end{array}
$$
7. First, in equation (1), replace $y$ with $f(y)$:
$$
\begin{array}{l} 
f(f(x)+f(y)) \\
= f(f(x)) + 2 f(y) f(x) - f(f(y)) + \\
2 f^{2}(y) + 1 \\
\Rightarrow f(f(x)+f(y)) - f(f(x)) - \\
2 f(y) f(x) - f(f(y)) \\
= -2 f(f(y)) + 2 f^{2}(y) + 1 .
\end{array}
$$

Notice that the left side of the equation is symmetric with respect to $x$ and $y$.
Therefore, the right side must also be symmetric with respect to $x$ and $y$.
Thus, $f(f(y)) = f^{2}(y) + c$, where $c$ is some constant to be determined.
Substituting into equation (2) gives:
$$
f(f(x) + f(y)) = (f(x) + f(y))^{2} + 1.
$$

Fix a positive constant $x_{0}$, and let $y_{0} = f(x_{0})$. Substituting into equation (1) gives:
$$
\begin{array}{l}
f(f(x_{0}) + y) \\
= f(f(x_{0})) + 2 y f(x_{0}) - f(y) + 2 y^{2} + 1 \\
\Rightarrow f(y_{0} + y) + f(y) \\
= f(y_{0}) + 2 y y_{0} + 2 y^{2} + 1.
\end{array}
$$

For any $t > f(y_{0}) + 1$, there exists $y_{t}$ such that:
$$
f(y_{0}) + 2 y_{t} y_{0} + 2 y_{t}^{2} + 1 = t.
$$

Thus, $f(y_{0} + y_{t}) + f(y_{t}) = t$.
Let $x = y_{0} + y_{t}$ and $y = y_{t}$, substituting into equation (3) gives:
$$
f(t) = t^{2} + 1 \quad (t > f(y_{0}) + 1).
$$

For any $x > f(y_{0}) + 1$, we have:
$$
x^{2} + 1 \geq 2 x > x > f(y_{0}) + 1.
$$

Substituting into equation (1) gives:
$$
\begin{array}{l}
f(x^{2} + 1 + y) \\
= f(x^{2} + 1) + 2 y (x^{2} + 1) - f(y) + 2 y^{2} + 1 \\
\text{By } f(x^{2} + 1 + y) = (x^{2} + 1 + y)^{2} + 1 \text{ and } \\
f(x^{2} + 1) = (x^{2} + 1)^{2} + 1
\end{array}
$$

Expanding this, we get:
$$
f(y) = y^{2} + 1.
$$

Substituting $f(x) = x^{2} + 1$ into equation (1) confirms that it holds.
Therefore, the solution to this problem is $f(x) = x^{2} + 1$.