I claim that the only solutions are $(x,n)=(6,14),(2,4), (5,10)$. We can easily verify that these are indeed solutions. Clearly, $x=1$ leads to no solution. Now assume $x\ge 2$ and let $m=\frac{n}{2}$. After dividing the equation by $4$, we get $3\cdot 2^{x-2}+1=m^2$, or $3\cdot 2^{x-2}=(m+1)(m-1)$. This leads to two cases. Case 1: $m+1=2^a$ and $m-1=3\cdot 2^b$ for nonnegative $a,b$ Then $2^a-3\cdot 2^b = 2$. We see that $a,b\ge 1$. Since $v_2(2)=1$, we need at least one of $a,b$ to be $1$. The case of $a=1$ leads to no solutions, but $b=1$ leads to the solution $(a,b)=(3,1)$, which corresponds to $m=7$, $n=14$, and $x=6$. Case 2: $m+1=3\cdot 2^a$ and $m-1=2^b$ for nonnegative $a,b$ Then $3\cdot 2^a-2^b = 2$. $a=b=0$ leads to $m = 2$, $n=4$, and $x=2$. Now assuming that $a,b>0$, we require $a=1$ or $b=1$. The case of $a=1$ leads to the solution $(a,b)=(1,2)$, which corresponds to $m=5$, $n=10$, and $x=5$. The case of $b=1$ leads to no solutions. We have exhausted all cases, so we are done.