What a technical problem. The answers are $\boxed{n=1 , 7}$. We claim these are the only ones. Suppose $n>1$ throughout the proof. Write $n^4-1=6m^2$. Since $(n^2-1)(n^2+1)=6m^2$, by mod 2, 3 reasons, it follows that $n^2+1= 2a^2, n^2-1=3(2b)^2$ for some $a, b\in \mathbb{N}$. Now, from $(n+1)(n-1)=12b^2$, or $\left(\frac{n+1}{2}\right)\left(\frac{n-1}{2}\right)=3b^2$, the only case is that $n+1=2p^2, n-1=6q^2$ for some $p, q\in \mathbb{N}$. Then $(a+1)(a-1)=6b^2=6p^2q^2,$ so we have four cases: Case 1. $a+1=4X^2, a-1=6Y^2$, $pq=2XY$ for some $X, Y \in \mathbb{N}$. In this case, we have $2X^2=3Y^2+1 \implies X^2 \equiv 2 \pmod 3$, which is impossible. Case 2. $a+1=12X^2, a-1=2Y^2$, $pq=2XY$ for some $X, Y \in \mathbb{N}$. In this case, we have $6X^2=Y^2+1 \implies Y^2 \equiv 2 \pmod 3$, which is impossible. Now here comes the hard cases. Case 3. $a+1=6X^2, a-1=4Y^2$, $pq=2XY$ for some $X, Y \in \mathbb{N}$. Since $p^2=3q^2+1$, $p$ and $q$ has different parity. We divide into two cases according to the parity of $p$: Subcase 1. $p$ is odd. Note that $p\times \frac{q}{2}=XY$, so there are integers $x, y, z, w$ such that $p=xy, q=2zw, X=xz, Y=yw$. Then we have the system of equations $$\begin{cases} 3X^2=2Y^2+1 \implies 3x^2z^2=2y^2w^2+1 \\ p^2=3q^2+1 \implies x^2y^2=12z^2w^2+1. \end{cases}$$Hence $3z^2(x^2+4w^2)=y^2(x^2+2w^2)$, then $\frac{x^2+4w^2}{x^2+2w^2}=\frac{y^2}{3z^2}$ are both reduced fractions(from $p^2=3q^2+1$), so $y^2=x^2+4w^2, 3z^2=x^2+2w^2$. Plugging back at $3x^2z^2=2y^2w^2+1$, we have $x^4=8w^4+1$. Then $\left(\frac{x^2-1}{4}\right)\left(\frac{x^2+1}{2}\right)=w^4$, so $x^2-1=4A^4$ for some positive integer $A$. Note that two positive squares aren't consecutive, contradiction. Subcase 2. $p$ is even. Following similarly(instead writing $p=2xy$ this time), we have \[ \begin{cases} 3X^2=2Y^2+1 \implies 3x^2z^2=2y^2w^2+1 \\ p^2=3q^2+1 \implies 4x^2y^2=3z^2w^2+1. \end{cases} \]Then $\frac{x^2+w^2}{2x^2+w^2}=\frac{2y^2}{3z^2}$ are both reduced(easy to check), so $2y^2=x^2+w^2, 3z^2=2x^2+w^2$. Now plugging back, we have $2x^4=w^4+1$. Then, \[ 4x^8=(w^4+1)^2=(w^4-1)^2+4w^4 \implies (x^2)^4-w^4=\left(\frac{w^4-1}{2}\right)^2, \]which is impossible if $w>1$. Hence the only case is $w=1 \implies x=1 \implies y=z=1 \implies a=5 \implies \boxed{n=7}$. Case 4. $a+1=2X^2, a-1=12Y^2$, $pq=2XY$ for some $X, Y \in \mathbb{N}$. The rest is similar with the first subcase above, so I'll just omit the unnecessary calculations. Subcase 1. $p$ is odd. Similarly we have $$\begin{cases} x^2z^2=6y^2w^2+1 \\ x^2y^2=12z^2w^2+1. \end{cases}$$Then $\frac{x^2+12w^2}{x^2+6w^2}=\frac{z^2}{y^2}$ are both reduced fractions, so $z^2=x^2+12w^2, y^2=x^2+6w^2$. Plugging back, we have $x^4=72w^4+1$. Then $\left(\frac{x^2-1}{36}\right)\left(\frac{x^2+1}{2}\right)=w^4$, so $x^2-1=36A^4=(6A^2)^2$ for some positive integer $A$, contradiction. Subcase 2. $p$ is even. Similarly we have $$\begin{cases} x^2z^2=6y^2w^2+1 \\ 4x^2y^2=3z^2w^2+1. \end{cases}$$Then $\frac{x^2+3w^2}{2x^2+3w^2}=\frac{2z^2}{y^2}$ are both reduced fractions, so $2z^2=x^2+3w^2, y^2=x^2+3w^2$. Since $q$ is odd, $w$ is odd too, so $x$ is odd. Then $z^2 \equiv 2 \pmod 4$, which is impossible. Now that we have exhausted all cases, the proof is complete. $\blacksquare$

Wow what a nice solution lminsl!

Related. http://matwbn.icm.edu.pl/ksiazki/aa/aa78/aa7847.pdf

Wait, why n²+1=2*a² I mean I understand why it is a multiple of 2 but why is it two times a square. Similarly why n²-1=3*(2b)²

ab=n^2, gcd(a,b)=1 then a=x^2, b=y^2 using this

@above Iminsl, what an awesome solution! I will present a solution using Pell's equations.

Lemma (1) There's no positive integer solution of $x^4 - 2y^2 = 1$. pf.

(2) There's no positive integer solution of $x^4 - 2y^2 =-1$ except $(x,y)=(1,1)$. pf.

$n^4 -1=6m^2 \implies (n^2)^2 - 6m^2 = 1$. (Note that $n$ is odd.) Fundamental solution of $u^2 -6v^2 = 1$ is $5+2\sqrt{6}$, therefore a solution satisfies $$u_k + v_k\sqrt{6} = (5+2\sqrt{6})^{k}$$Note that $u_0 = 1, u_1 = 5, u_{k+1} = 10u_k - u_{k-1}, u_k = \frac{1}{2} \left((5+2\sqrt{6})^k + (5-2\sqrt{6})^k \right)$, $u_k$ is odd. Now, we have $n^2=u_k$ for some $k\geq 0$. Case 1. If $k$ is odd, then $u_k\equiv -1(\text{mod} 3)$. Contradiction! Case 2. If $4|k$, let $k=4l$. $u_{4l} = 2u_{2l}^2 - 1 = 2(2u_l^2-1)^2-1 = 8u_l^4 - 8u_l^2 +1 = (4u_l^2 -1)^2 - 8u_l^4$ $\implies 8u_l^4 = (4u_l^2 -1)^2 - n^2 = (4u_l^2 -1-n)(4u_l^2 -1+n)$. $\gcd(4u_l^2-1,8u_l^4)=1\implies\gcd(4u_l^2-1-n,4u_l^2-1+n)=2$. There exist odd $a,b$ s.t. $u_l=ab,\{4u_l^2-1-n, 4u_l^2-1+n\} = \{4a^4, 2b^4\} \implies 4(ab)^2-1 = 4u_l^2-1 = 2a^4 + b^4 \implies b^4 - 2(a^2-b^2)^2 = 1$. By Lemma, $a=b=1$. So, $n = u_0 = 1$. Case 3. If $2|k$ and $4\nmid k$, let $k=4l+2$. Note that $5+2\sqrt{6} = \frac{1}{2} (2+\sqrt{6})^2$. $$u_{2l+1} = \frac{1}{2}\left((5+2\sqrt{6})^{2l+1} + (5-2\sqrt{6})^{2l+1} \right) = \frac{1}{2^{2l+2}} \left( (2+\sqrt{6})^{2l+1}+(2-\sqrt{6})^{2l+1} \right)^2 + 1 = \left(\frac{ (2+\sqrt{6})^{2l+1}+(2-\sqrt{6})^{2l+1}}{2^{l+1}} \right)^2 + 1$$Hence, $u_{2l+1} =x^2 +1$ for some positive integer $x$. $n^2 = u_{4l+2} = 2u_{2l+1}^2 -1 = 2(x^2 +1)^2 -1 \implies 2x^4 = (2x^2 +1)^2-n^2 = (2x^2+1-n)(2x^2+1+n)$. $\gcd(2x^2+1,2x^4)=1\implies\gcd(2x^2+1-n,2x^2+1+n)=2$. There exists positive integers $a,b$ s.t. $x=2ab$, $\{2x^2+1-n, 2x^2+1+n\}=\{16a^4,2b^4\} \implies 2(2ab)^2+1=2x^2+1=8a^4 +b^4 \implies b^4 - 2(2a^2-b^2)^2=-1$. By Lemma, $a=b=1$. Therefore, $n^2 = u_2 = 49\implies n=7$. Answer: $\boxed{n=1,7}$

$n^4-1=6u^2 \rightarrow (n^2-1)(n^2+1)=6u^2$ we can easily see $n^2-1=12a^2 ;n^2+1=2b^2$ from this we can find this $6a^2+1=b^2$ and $6a^2=(b-1)(b+1)$ let $a^2=2^{2k}\cdot c^2 \cdot d^2 \rightarrow 2^{2k+1} \cdot 3 \cdot c^2 \cdot d^2=(b-1)(b+1)$ we have four posible option $b-1=c^22^{2k}3$;$b+1=2d^2$ $\rightarrow$$2^{2k-1}c^23+1=d^2$ $b-1=c^22^{2k}$;$b+1=6d^2$ $\rightarrow$ $1+c^22^{2k-1}=3d^2$ by modulo $4$,$k=1$ and we famous problem with answers $c,d=1$ $b-1=c^26$;$b+1=d^22^{2k}$ $\rightarrow$ $3c^2+1=2^{2k-1}d^2$ by modulo $3$ its impossible $b-1=2c^2$;$b+1=d^22^{2k}3$ $\rightarrow$ $c^2+1=2^{2k-1}d^23$ by modulo $3$ impossible so we have only one case $2^{2k-1}c^23+1=d^2$ $c$ is odd so let $c^2=m^2l^2$ and we will have again $2^{2k-1}l^2m^23=(d-1)(d+1)$ and we will have endless descent until $k=1$ where $c$ will be not found and we dont have solution in this case or until we will find "famous problem" and again unfinied so we have solution only $n=7,1$