I did try to search for a discussion of this problem before but I could find nothing! Here is a solution nonetheless..
If $a=1$, then $2^a+3^b=n^2=2\pmod{3}$, which is impossible. So $a\ge 2$ so $3^b=n^2\pmod{4}$ meaning $b$ is even, let it be $2c$. Then $2^a=(n-3^c)(n+3^c)$. Let $n-3^c=2^r$ and $n+3^c=2^s$ where $s>r$. Then $2^s-2^r=2\cdot 3^c$ i.e. $2^{s-1}-2^{r-1}=3^c$. This means the $LHS$ is odd, which can only happen when $r-1=0$ so $r=1$ meaning $n=3^c+2$. Substituting this back into the original equation yields $2^a=4\cdot 3^c+4$. Thus $2^a||4(3^c+1)$. We now examine the largest power of $2$ that can divide $3^c+1$. Looking $\pmod{8}$, the expression can equal either $4$ or $2$. Thus it can only contribute at most $2$ to the value of $a$, i.e. $2\le a\le 4$. If $a=4$ then $2^4=4\cdot 4^c+1$ yielding the solution $(4,2)$. The case $a=2$ reaches a contradiction, the case $a=3$ gives the invalid solution $(3,0)$.