Here is a solution using less machinery than in Will Jagy's answer.
As Daniel Robert-Nicoud noted in comments, the fact that $3^a+7^b\equiv(-1)^a+(-1)^b$ mod $4$ implies $a$ and $b$ must have opposite parity in order for the (even) sum $3^a+7^b$ to be a perfect square. So we seek to show that $(a,b)=(1,0)$ is the only solution in the (odd,even) case and $(a,b)=(2,1)$ is the only solution in the (even,odd) case.
If $a$ is odd and $b=2m$ is even, then we can rewrite $3^a+7^b=n^2$ as $3^a=(n+7^m)(n-7^m)$. This implies $n+7^m$ and $n-7^m$ are each powers of $3$, say $3^c$ and $3^d$ with $c+d=a$. But then $2\cdot7^m=3^c-3^d$, so we must have $d=0$ and $c=a$, since $3\not\mid2\cdot7^m$. One solution to $2\cdot7^m=3^a-1$ is $(a,m)=(1,0)$. It remains to show there are no solutions with $m\gt0$. For this it suffices to note that $3^a\equiv1$ mod $7$ if and only if $6\mid a$, which contradicts the assumption that $a$ is odd.
In the other parity case, if $a=2m$ is even and $b$ is odd, then, along similar lines, we must have $2\cdot3^m=7^b-1$, which has $(m,b)=(1,1)$ as one solution. After checking that there is no solution with $m=0$, it remains to show there are no solutions with $m\gt1$. If there were, then we would have $7^b\equiv1$ mod $9$, which would imply $b\equiv3$ mod $6$ (since $6$ is the order of the multiplicative group of units mod $9$ and $7^3\equiv(-2)^3=-8\equiv1$ mod $9$). Writing $b=3r$ (with $r$ odd, but that doesn't matter), we note that $(7^3-1)\mid(7^{3r}-1)$, so that $7^3-1=2\cdot3^2\cdot19$ divides $2\cdot3^m$, an obvious contradiction.
