If $x = 1$, then we have no solutions, while for $x = 2$, we have the solution $(x, n) = (2, 4)$.

Now we suppose that $x \ge 3$. The left part of the equation is even and so $n$ is even, say $n = 2k$. Then the equation can be written as: $3 \cdot 2^x + 4 = 4k^2 \Leftrightarrow 3 \cdot 2^{x-2} = k^2 - 1$, and hence $k$ is odd. Thus the equation is written as: $3 \cdot 2^{x-2} = (k-1)(k+1)$.

1st case: $k+1 = 3 \cdot 2^a$ and $k-1 = 2^b$, $a, b \ge 1$, with $a+b = x-2$. Then by subtracting we get $3 \cdot 2^a - 2^b = 2$. If $a, b \ge 2$, then the left part is divisible by 4, while the right part is not. Therefore, either $a=1$, which implies $b=2$ and $x=5$, or $b=1$, not giving solution for $a$. Hence we have $(x, n) = (5, 10)$.

2nd case: $k+1 = 2^s$ and $k-1 = 3 \cdot 2^t$, $s, t \ge 1$, with $s+t = x-2$. Then we have the equation $2^s - 3 \cdot 2^t = 2$. If $s, t \ge 2$ then the left part is divisible by 4, while the right part is not. Hence, either $s=1$ or $t=1$, giving $s=3$ and $x=6$. Therefore in this case we have the solution $(x, n) = (6, 14)$.