Find all $n,x \in \Bbb N$ such that $3\cdot2^x+4=n^2$ 
Find all $n,x \in \Bbb N$ such that $3\cdot2^x+4=n^2$

Arranging the equation a bit one has that $$(n-2)(n+2)=3\cdot2^x.$$
Now considering the cases $n-2=3, n+2=2^x$ from $n-2=3 \Rightarrow n=5$. Substituting this to $n+2=2^x \Rightarrow 7=2^x$ which isn't an integer solution.
Considering the opposite $n+2=3$ and $n-2=2^x$ one gets that $n+2=3 \Rightarrow n=1$ and substituting this to $n-2=2^x$ we have that $-1=2^x$ which also doesn't have any solutions.
What am I missing here since this seemed to have $3$ pairs of solutions?

You're off to a good start; writing
$$(n-2)(n+2)=3\cdot2^x,$$
shows, by unique factorization of integers, that
$$n-2=2^a3^b,\qquad\text{ and }\qquad n+2=2^c3^d,$$
for nonnegative integers $a$, $b$, $c$ and $d$ with $a+c=x$ and $b+d=1$. Of course $b+d=1$ leaves very few options for $b$ and $d$. Can you continue from here?
Solution:

 It follows that $$2^{x-a}3^{1-b}-2^a3^b=(n+2)-(n-2)=4,$$ so if $x\geq3$ we see that either $a\geq2$ or $x-a\geq2$. Then reducing mod $4$ shows that in fact  $a\geq2$ and $x-a\geq2$, and so $$2^{x-a-2}3^{1-b}-2^{a-2}3^b=(n+2)-(n-2)=1.$$  Then precisely one of the two factors on the left hand side is odd, meaning that either $x-a-2=0$ or $a-2=0$, and so correspondingly either $$3^{1-b}-2^{a-2}3^b=1\qquad\text{ or }\qquad2^{x-4}3^{1-b}-3^b=1.$$ For the former, note that $3^{1-b}\in\{1,3\}$. Clearly $3^{1-b}=1$ does not yield a solution, so $3^{1-b}=3$ and hence $b=0$. Then $2^{a-2}=2$ and so $a=3$, yielding the solution $x=5$ with $n=10$.  For the latter, note again that $3^b\in\{1,3\}$. A quick check shows that $3^b=1$ does not yield a solution, so $3^b=3$ and hence $b=1$. Then $2^{x-4}=4$, yielding the solution $x=6$ with $n=14$.   Finally we are left with the case $x\leq3$. Another quick check yields the last solution $x=2$ with $n=4$.

