Solution:

The answer is 4040. Let $a$ be the leading coefficient of $p(x)$; comparing the leading coefficients of $p(x)^2$ and $p\left(x^2\right)+k p(x)$ we get $a^2=a$, hence $a=1$, that is, $p(x)$ is monic. Now, if $p(x)$ is a monomial we always have $p(x)^2=p\left(x^2\right)$, that is, the desired equality holds with $k=0$. Otherwise $p(x)=x^n+r(x)$ with $r(x)$ a nonzero polynomial of degree $m<n$. Then $p(x)^2$ is $x^{2 n}$ plus a polynomial of degree $n+m$, while $p\left(x^2\right)+k p(x)$ is $x^{2 n}$ plus a polynomial of degree at most $\max \{2 m, n\}$. The only possibility is that $m=0$ and hence $p(x)=x^n+b$ with $b \in \mathbb{R} \setminus \{0\}$. Substituting into the equation we get $x^{2 n}+2 b x^n+b^2=x^{2 n}+k x^n+(k+1) b$, from which $k=2 b$ and hence $b^2=(2 b+1) b$, that is, $b=-1$. The solutions are therefore all and only those of the form $x^n$ or $x^n-1$, for a total of 4040 polynomials.