We will show that $f(x) = x$ for all $x$. Substituting $0$ for $x$ in $*$ we obtain
$$(1) \qquad f(f(y) + 1) = y + (f(1))^2.$$
Then substituting $0$ for $x$ and $f(y) + 1$ for $y$ in $*$ and using (1) we get
$$f(y + (f(1))^2 + 1) = f(y) + 1 + (f(1))^2$$
and now substituting $x^2 + y$ for $y$ gives
$$f(x^2 + y + (f(1))^2 + 1) = f(x^2 + y) + 1 + (f(1))^2.$$
Now using (1) and $*$ for the left hand side we get
$$f(y) + 1 + (f(1+x))^2 - 2x = f(x^2 + y) + 1 + (f(1))^2$$
and hence
$$(2) \qquad f(x^2 + y) - f(y) = (f(x+1))^2 - 2x - (f(1))^2.$$
In particular,
$$(3) \qquad f(x^2 + y) - f(y) = f(x^2) - f(0).$$
Now letting $y = 0$ and $x = 1$, $y = 0$ and $x = -1$, and $y = 1$ and $x = 1$ in (2) we obtain $(f(2))^2 - 2 - (f(1))^2 = f(1) - f(0) = (f(0))^2 + 2 - (f(1))^2$ and $f(2) = 2f(1) - f(0)$.

$$(4) \qquad f(x^2 + y) = f(x^2) + f(y)$$
and hence
$$(5) \qquad f(x+y) = f(x) + f(y)$$
for all $x$ and $y$. Combining (4) with (1) we also get
$$f(f(y)) = y$$
for all $y$, and using this and (5) we obtain
$$(6) \qquad f(x^2) + 2x = (f(x))^2 + 2f(x)$$
from $*$. Now using the equations obtained by substituting $y + f(x)$, $f(x)$ and $y$ for $x$ in (6) gives $f(2yf(x)) = 2xf(y)$ for all $x$ and $y$. Using (5) we get $f(yf(x)) = xf(y)$, and hence $f(xy) = f(x)f(y)$ for all $x$ and $y$. In particular, $f(x^2) = f(x)^2$ for all $x$ and now (6) gives $f(x) = x$ for all $x$.