First solution. Since the product of $n$ consecutive integers is divisible by $n!$, it is sufficient to show that there are infinitely many positive integers $n$ such that $n!$ is divisible by $(n+k)^2 + 1$.
To obtain infinitely many positive integers $n$ such that $n!$ is divisible by $(n+k)^2 + 1$, it is sufficient to consider large enough integers $m \equiv 1 \pmod{5}$ and let $n = 2m^2 - k$.

In this case, $(n + k)^2 + 1 = 4m^4 + 1 = (2m^2 + 2m + 1)(2m^2 - 2m + 1) = 5\ell(2m^2 - 2m + 1)$, and $5 < \ell < 2m^2 - 2m + 1 < 2m^2 - k = n$, so $n!$ is indeed divisible by $(n + k)^2 + 1$.

Second solution. To obtain infinitely many positive integers $n$ such that $(n+k)^2 + 1$ divides $n!$, consider all pairs $(n, m)$ of positive integers satisfying $n^2 - 5m^2 = -1$. These pairs are completely described by
$$\begin{pmatrix} n_r \\ m_r \end{pmatrix} = \begin{pmatrix} 9 & 20 \\ 4 & 9 \end{pmatrix}^r \begin{pmatrix} 2 \\ 1 \end{pmatrix}, \quad r = 0, 1, 2, \dots;$$
the $n_r$ and the $m_r$ both form strictly increasing sequences of positive integers.
Write $2(n_r^2 + 1) = 5 \cdot m_r \cdot 2m_r$ and notice that $m_r < 2m_r \le \sqrt{5m_r^2 - 1} - k = n_r - k$ for all but finitely many indices, to conclude that $(n_r - k)!$ is divisible by $n_r^2 + 1$ for all but finitely many indices.