Counting them is difficult!

Since the expression is a perfect square so we can note down it as $\begin{aligned} &3^{2n} +3n^2 +7 = d^2 \\& d^2 = (3^n)^{2}+3n^2 +7 \\& 3n^2 +7 = (d-3^n)(d+3^n)\end{aligned}$ Shows that the left part it is product of two factors also it cannot have factors more than 2 . If the prime factor of $3n^2+7$ are less than or equal 7 then the integer $n$ doesn't exist.

So it must only be a prime number such that $n^2$ is pefect square number. That is $\begin{aligned} &3n^2 +7 = (d-3^n)(d+3^n) = 1 \times (d+3^n) \\& 3n^2+7 = d+3^n \implies n = \sqrt{\dfrac{d+3^n-7}{3}}\end{aligned}$ To yield an integer $n$ , $d+3^n$ must be 19. $\begin{aligned}\therefore n = \sqrt{\dfrac{19-7}{3}} = 2 \quad 3n^2+7 = 1\phantom{c}, n = \sqrt{-2} \end{aligned}$ Only the acceptable value for $n$ is $2$ . So number of solutions is only $1$ .