Solved with Enigma714. Easy cases $n=1,2,4$, suppose $n \geq 5$ Lemma: $n= 2^k$ Suppose $d>1$ is an odd divisor of $n$ then $3^d-2^d \mid 3^n-2^n=x^2+1$ which its odd divisors are $\equiv 1 \pmod{4}$ but $3^d-2^d$ is not, our lemma follows. Using LTE: $v_{2}(3^n-1)=v_{2}(3-1)+v_2(3+1) + v_{2}(n) - 1 = k+2$ and since $v_{2}(2^{n})=n > k +2$ we get that $v_{2}(x^2)= k+2$. Notice that: $(3^{n/2}-x)(3^{n/2}+x)=2^{2^k}+1$ and is well-known that all divisors of RHS are $\equiv 1 \pmod{2^{k+1}}$ then: $$(3^{n/2}+x)-(3^{n/2}-x) \equiv 0 \pmod{2^{k+1}} \Rightarrow x \equiv 0 \pmod{2^k}$$Which give us a contradiction with $v_{2}(x^2)= k+2$. So all the possible values of $n$ are $1,2,4$.

$\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ Suppose there is a solution with $n\geq 5$ $3^n-2^n-1=x^2\Rightarrow 2|x^2 \Rightarrow 4|x^2=3^n-2^n-1\Rightarrow 4|3^n-1\Rightarrow 4|(-1)^n-1\Rightarrow 2|n$, let $k=v_2(n)\geq 1$ and $n=2^kt \Rightarrow (3^t-2^t)(3^t+2^t)(3^{2t}+2^{2t})\ldots (3^{2^{k-1}t}+2^{2^{k-1}t})=3^{2^kt}-2^{2^kt}=x^2+1\Rightarrow \forall p \text{ prime }/p|3^t-2^t, p\equiv 1\pmod{4}, t>1$ $\Rightarrow 3^t-2^t\equiv 1 \pmod{4} \Rightarrow 4|(-1)^t-1\Rightarrow 2|t, t>1(\Rightarrow\Leftarrow)$ $\Rightarrow t=1\Rightarrow n=2^k\geq 5\Rightarrow k\geq 3\Rightarrow 3^{2^k}-2^{2^k}-1=x^2$ Lemma: $(3^{2^{k-1}}-2)^2<3^{2^k}-2^{2^k}-1,\forall k\geq 3$ Proof: When I return to my country I will write it because I am lazy but it is only an induction $\Rightarrow (3^{2^{k-1}}-2)^2<x^2<3^{2^k}\Rightarrow x=3^{2^{k-1}}-1$ $\Rightarrow 3^{2^k}-2^{2^k}-1=(3^{2^{k-1}}-1)^2\Rightarrow 2.3^{2^{k-1}}=2+2^{2^k}\Rightarrow 3^{2^{k-1}}=1+2^{2^k-1}$ By Mihailescu's Theorem $(\Rightarrow \Leftarrow)$ $\Rightarrow n=1,2,4$ are the only solutions$_\blacksquare$ $\color{blue}\rule{24cm}{0.3pt}$

$\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ Suppose there is a solution with $n\geq 5$ $3^n-2^n-1=x^2\Rightarrow 2|x^2 \Rightarrow 4|x^2=3^n-2^n-1\Rightarrow 4|3^n-1\Rightarrow 4|(-1)^n-1\Rightarrow 2|n$, let $k=v_2(n)\geq 1$ and $n=2^kt \Rightarrow (3^t-2^t)(3^t+2^t)(3^{2t}+2^{2t})\ldots (3^{2^{k-1}t}+2^{2^{k-1}t})=3^{2^kt}-2^{2^kt}=x^2+1\Rightarrow \forall p \text{ prime }/p|3^t-2^t, p\equiv 1\pmod{4}, t>1$ $\Rightarrow 3^t-2^t\equiv 1 \pmod{4} \Rightarrow 4|(-1)^t-1\Rightarrow 2|t, t>1(\Rightarrow\Leftarrow)$ $\Rightarrow t=1\Rightarrow n=2^k\geq 5\Rightarrow k\geq 3\Rightarrow 3^{2^k}-2^{2^k}-1=x^2$ Lemma: $(3^{2^{k-1}}-2)^2<3^{2^k}-2^{2^k}-1,\forall k\geq 3$ Proof: When I return to my country I will write it because I am lazy but it is only an induction $\Rightarrow (3^{2^{k-1}}-2)^2<x^2<3^{2^k}\Rightarrow x=3^{2^{k-1}}-1$ $\Rightarrow 3^{2^k}-2^{2^k}-1=(3^{2^{k-1}}-1)^2\Rightarrow 2.3^{2^{k-1}}=2+2^{2^k}\Rightarrow 3^{2^{k-1}}=1+2^{2^k-1}$ By Mihailescu's Theorem $(\Rightarrow \Leftarrow)$ $\Rightarrow n=1,2,4$ are the only solutions$_\blacksquare$ $\color{blue}\rule{24cm}{0.3pt}$

$\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ Suppose there is a solution with $n\geq 5$ $3^n-2^n-1=x^2\Rightarrow 2|x^2 \Rightarrow 4|x^2=3^n-2^n-1\Rightarrow 4|3^n-1\Rightarrow 4|(-1)^n-1\Rightarrow 2|n$, let $k=v_2(n)\geq 1$ and $n=2^kt \Rightarrow (3^t-2^t)(3^t+2^t)(3^{2t}+2^{2t})\ldots (3^{2^{k-1}t}+2^{2^{k-1}t})=3^{2^kt}-2^{2^kt}=x^2+1\Rightarrow \forall p \text{ prime }/p|3^t-2^t, p\equiv 1\pmod{4}, t>1$ $\Rightarrow 3^t-2^t\equiv 1 \pmod{4} \Rightarrow 4|(-1)^t-1\Rightarrow 2|t, t>1(\Rightarrow\Leftarrow)$ $\Rightarrow t=1\Rightarrow n=2^k\geq 5\Rightarrow k\geq 3\Rightarrow 3^{2^k}-2^{2^k}-1=x^2$ Lemma: $(3^{2^{k-1}}-2)^2<3^{2^k}-2^{2^k}-1,\forall k\geq 3$ Proof: When I return to my country I will write it because I am lazy but it is only an induction $\Rightarrow (3^{2^{k-1}}-2)^2<x^2<3^{2^k}\Rightarrow x=3^{2^{k-1}}-1$ $\Rightarrow 3^{2^k}-2^{2^k}-1=(3^{2^{k-1}}-1)^2\Rightarrow 2.3^{2^{k-1}}=2+2^{2^k}\Rightarrow 3^{2^{k-1}}=1+2^{2^k-1}$ By Mihailescu's Theorem $(\Rightarrow \Leftarrow)$ $\Rightarrow n=1,2,4$ are the only solutions$_\blacksquare$ $\color{blue}\rule{24cm}{0.3pt}$

mijail's solution proved $n=2^k$, so I'm only showing how I finished: Claim:For $k \ge 1$, $2^{k+2}$ is the largest power of $2$ that divides $3^{2^k}-1$. Proof: Let $x_k=\dfrac{3^{2^k}-1}{2^{k+2}}$. We shall prove that $x_k$ is an odd integer. Case $x_1=1$ is clear. Next, note that $x_{k+1}=\left(\dfrac{3^{2^k}+1}{2}\right)\left(\dfrac{3^{2^k}-1}{2^{k+2}}\right)=\left(\dfrac{3^{2^k}+1}{2}\right)x_k$, the product of $x_k$ by an odd integer, since $3^{2^k}+1 \equiv 2 \pmod{4}$. For $k=0,1,2$, we have the solutions $\boxed{n=1,2,4}$. Assume $k>2$. We have $3^n-2^n-1=2^{k+2}x_k-2^{2^k}=2^{k+2}(x_k-2^{2^k-k-2})$. This implies that $2^{k+2}$ is a perfect square, since $x_k-2^{2^k-k-2}$ is odd: indeed, the exponent $2^k-k-2$ is at least $3$, because $2^k=(1+1)^k>1+k+\binom{k}{2} \ge 1+k+\binom{3}{2} \Rightarrow 2^k \ge k+5$. Therefore, the factor $x_k-2^{2^k-k-2}$, besides being itself a perfect square, has to be equivalent to $x_k$ modulo $8$. This is impossible, because $5$ is a quadratic nonresidue modulo $8$, and $x_k \equiv 5 \pmod{8}$ para $k \ge 2$. To verify this, note that $x_2=5$ and, for $k \ge 2$, we have $3^{2^k} \equiv 1 \pmod{16}$. Thus $3^{2^k}+1=16l+2 \Rightarrow x_{k+1}=(8l+1)x_k \equiv x_k \pmod{8}$, completing the proof by induction.