The answer is $n = 3, 4, 5, 7$.
We first introduce a variant of the $k$th Chebyshev polynomials in the following lemma (which is standard, and easily shown by induction).

Lemma
For each $k \ge 0$ there exists $P_k(X) \in \mathbb{Z}[X]$, monic for $k \ge 1$ and with degree $k$, such that
$$P_k(X + X^{-1}) \equiv X^k + X^{-k}.$$
The first few are $P_0(X) \equiv 2$, $P_1(X) \equiv X$, $P_2(X) \equiv X^2 - 2$, $P_3(X) \equiv X^3 - 3X$.

Suppose the angles of the triangle are $\alpha < \beta < \gamma$, so the law of cosines implies that
$$2 \cos \alpha = \frac{n+4}{n+1} \quad \text{and} \quad 2 \cos \gamma = \frac{n-4}{n-1}.$$

Claim — The triangle is quirky iff there exists $r, s \in \mathbb{Z}_{\ge 0}$ not both zero such that
$$\cos(r\alpha) = \pm \cos(s\gamma) \quad \text{or equivalently} \quad P_r\left(\frac{n+4}{n+1}\right) = \pm P_s\left(\frac{n-4}{n-1}\right).$$
*Proof.* If there are integers $x, y, z$ for which $x\alpha + y\beta + z\gamma = 0$, then we have that $(x - y)\alpha = (y - z)\gamma - \pi y$, whence it follows that we may take $r = |x - y|$ and $s = |y - z|$ (noting $r = s = 0$ implies the absurd $x = y = z$). Conversely, given such $r$ and $s$ with $\cos(r\alpha) = \pm \cos(s\gamma)$, then it follows that $r\alpha \pm s\gamma = k\pi = k(\alpha + \beta + \gamma)$ for some $k$, so the triangle is quirky. $\square$

If $r = 0$, then by rational root theorem on $P_s(X) \pm 2$ it follows $\frac{n-4}{n-1}$ must be an integer which occurs only when $n = 4$ (recall $n \ge 3$). Similarly we may discard the case $s = 0$.

Thus in what follows assume $n \ne 4$ and $r, s > 0$. Then, from the fact that $P_r$ and $P_s$ are nonconstant monic polynomials, we find

Corollary
If $n \ne 4$ works, then when $\frac{n+4}{n+1}$ and $\frac{n-4}{n-1}$ are written as fractions in lowest terms, the denominators have the same set of prime factors.

But $\gcd(n+1, n-1)$ divides 2, and $\gcd(n+4, n+1)$, $\gcd(n-4, n-1)$ divide 3. So we only have three possibilities:

* $n + 1 = 2^u$ and $n - 1 = 2^v$ for some $u, v \ge 0$. This is only possible if $n = 3$. Here $2 \cos \alpha = \frac{7}{4}$ and $2 \cos \gamma = -\frac{1}{2}$, and indeed $P_2(-1/2) = -7/4$.

* $n + 1 = 3 \cdot 2^u$ and $n - 1 = 2^v$ for some $u, v \ge 0$, which implies $n = 5$. Here $2 \cos \alpha = \frac{3}{2}$ and $2 \cos \gamma = \frac{1}{4}$, and indeed $P_2(3/2) = 1/4$.

* $n + 1 = 2^u$ and $n - 1 = 3 \cdot 2^v$ for some $u, v \ge 0$, which implies $n = 7$. Here $2 \cos \alpha = \frac{11}{8}$ and $2 \cos \gamma = \frac{1}{2}$, and indeed $P_3(1/2) = -11/8$.

Finally, $n = 4$ works because the triangle is right, completing the solution.