Suppose that $n$ is a solution to the problem. The polynomial
$$P(x) = x^n - 4x^2 + 2x + 1$$
clearly has a root at $1$. Therefore we may write $P(x) = (x - 1)Q(x)$ for some polynomial $Q$. Since $P(x) \ge 0$ for $x > 0$, the polynomial $Q$ changes sign at $1$ and so $Q(1) = 0$. Calculating
$$Q(x) = \frac{P(x)}{x-1} = \frac{x^n - 1}{x-1} - \frac{4x^2 - 2x - 2}{x-1} = (1 + x + \dots + x^{n-1}) - (4x + 2)$$
we see that $Q(1) = n - 6$ and hence $n = 6$.

Conversely, let $n = 6$. Note that $Q(x) = (x - 1)R(x)$, where $R$ is a polynomial given by:
$$R(x) = x^4 + 2x^3 + 3x^2 + 4x + 1$$
Since $P(x) = (x - 1)^2 R(x)$ and $R(x) \ge 0$ for $x > 0$, we indeed have that $P(x) \ge 0$ for $x > 0$. It follows that $n = 6$ is a solution, and thus the only solution, to the problem.

Let $n$ be a solution to the problem. Consider the function $P : (0, +\infty) \to \mathbb{R}$, defined by $P(x) = x^n - 4x^2 + 2x + 1$ for every $x > 0$. Since $P(x) \ge 0$ for every $x > 0$ and $P(1) = 0$, it follows that the function $P$ attains its minimum at $x = 1$. Since this is also a local minimum and $P$ is differentiable, we have $P'(1) = 0$. Therefore, $P'(1) = n - 8 + 2 = 0$. Hence, $n = 6$ is the only possible value that can be a solution to the problem. On the other hand, for $x > 0$, using the AM-GM inequality, we have
$$x^6 + x + x + 1 \ge 4\sqrt[4]{x^6 \cdot x \cdot x \cdot 1} = 4x^2$$
This proves that $n = 6$ is the only solution to the problem.

We shall prove that the only solution is $n = 6$.
The given condition is equivalent to $P(x) \ge 0$ for all $x > 0$, where $P(x) = x^n - 4x^2 + 2x + 1$. Using elementary transformations, for $n > 2$, we have:
$$\begin{align*} P(x) &= x^n - 1 - 4x^2 + 2x + 2 = (x-1)(x^{n-1} + x^{n-2} + \dots + x + 1) - (x-1)(4x+2) = \\ &= (x-1)(x^{n-1} + x^{n-2} + \dots + x^2 - 3x - 1) = (x-1)(x^{n-1} - 1 + x^{n-2} - 1 + \dots + x^2 - 1 - 3x + 3 + n - 6) = \\ &= (x-1)\left((x-1)(x^{n-2} + 2x^{n-3} + 3x^{n-4} + \dots + (n-2)x + n - 5) + n - 6\right) = \\ &= (x-1)^2(x^{n-2} + 2x^{n-3} + 3x^{n-4} + \dots + (n-2)x + n - 5) + (n-6)(x-1). \tag{P} \end{align*}$$
From (P), since for $n = 6$ and $x > 0$ we have $Q(x) > 0$, it follows that $n = 6$ is one solution. Since the polynomial $Q(x)$ has all positive coefficients (except possibly the constant term), we have $Q(1 - \frac{1}{3n}) < Q(1) = \frac{n^2-n-8}{2}$, and based on (P), for $n \ge 7$, it holds that
$$P\left(1 - \frac{1}{3n}\right) = \frac{Q\left(1 - \frac{1}{3n}\right)}{9n^2} - \frac{n-6}{3n} < \frac{Q(1)}{9n^2} - \frac{n-6}{3n} = \frac{5n(7-n) - 8}{18n^2} < 0.$$
This shows that the numbers $n \ge 7$ are not solutions to the problem. On the other hand, for $1 \le n \le 5$, we have
$$P(1.1) \le 1.1^5 - 4 \cdot 1.1^2 + 2 \cdot 1.1 + 1 = -0.02949 < 0,$$
so the numbers $1 \le n \le 5$ are also not solutions.

Let $P(x) = x^n + 2x + 1 - 4x^2$, we want to check if $P(x) \ge 0$ holds for all $x > 0$. If $n = 1$ substituting $x = 2$ yields $P(x) = -11 < 0$, meaning that $n = 1$ cannot be our solution. From now on, assume that $n \ge 2$, and substitute $x = 1 + t$.
Using the binomial theorem we obtain that $P(1+t) = (1+t)^n + 2(1+t) - 4(1+t)^2 = (n-6)t - 4t^2 + \sum_{k=2}^n \binom{n}{k} t^k \le (n-6)t + \sum_{k=2}^n \binom{n}{k} t^k$. It is trivially true for $k \ge 2$ that $\binom{n}{k} t^k < n^k |t|^k$, which implies $P(1+t) < (n-6)t + (nt)^2 \sum_{k=0}^{n-2} (n|t|)^k$. Utilizing the identity that for $x \ne 1$ we have that $\sum_{k=0}^m x^k = \frac{1-x^{m+1}}{1-x}$ we obtain that $P(1+t) < (n-6)t + (nt)^2 \frac{1-n^{n-1}|t|^{n-1}}{1-n|t|} = t(n-6 + tn^2 \frac{1-n^{n-1}|t|^{n-1}}{1-n|t|})$.
If $n < 6$, then let $t = \frac{1}{n^3}$, plugging this in we obtain $P\left(1 + \frac{1}{n^3}\right) < \frac{1}{n^3} \left(n - 6 + \frac{1-\frac{1}{n^2n-2}}{n-\frac{1}{n}}\right)$. Now since $n < 6$ and $n \in \mathbb{N}$, we have that $n - 6 \le -1$. On the other hand, we know that $1 - \frac{1}{n^2n-2} < 1 < \frac{3}{2} \le n - \frac{1}{n}$ for $n \ge 2$, implying that $\frac{1-\frac{1}{n^2n-2}}{n-\frac{1}{n}} < 1$. Therefore the expression in the parenthesis is negative, while $\frac{1}{n^3} > 0$, meaning that the whole expression is negative, or $P\left(1 + \frac{1}{n^3}\right) < 0$, meaning that we found a negative value for $P(x)$ for $x = 1 + \frac{1}{n^3} > 0$. Therefore no $n \in \{2, 3, 4, 5\}$ can be a solution.
If $n > 6$, then let $t = -\frac{1}{n^3}$. Similarly to the case $n < 6$, we obtain that $P(1 - \frac{1}{n^3}) < -\frac{1}{n^3} \left(n - 6 + \frac{1 - \frac{1}{n^2n-2}}{n - \frac{1}{n}}\right)$. Note that $n - 6 \ge 1$, and it still holds that $\frac{1 - \frac{1}{n^2n-2}}{n - \frac{1}{n}} < 1$ using the same arguments. Therefore, the expression in the parenthesis is positive, while $-\frac{1}{n^3} < 0$, meaning that $P\left(1 - \frac{1}{n^3}\right) < 0$, and since for $n > 6$ we have $1 - \frac{1}{n^3} > 0$, we again found a negative point of our polynomial for some positive input.
Therefore, the only case left is when $n = 6$. For this take $x^6 + x + x + 1 \ge 4x^2$ using AM-GM. Therefore, the only solution is $n = 6$.