EDIT: I'm using a lemma: the sum of two squareroots of natural numbers can be rational only when they both are in fact natural numbers (see here)
Let $a \geq b$ (due to symmetry we can assume this, then all solutions will be permutations of the ones we find).
Denote $q = \frac{b}{a}$. Now $0<q\leq 1$.
Let's complete a square under the first squareroot:
$$
a^2 + 6b
= a^2 + 2\cdot 3qa + (3q)^2 - (3q)^2
= (a+3q)^2 - 9q^2
$$
Since this must be a square and bounded below by $a^2+1$ (since $b>0$) and from above (stricly since $q>0$) by $(a+3)^2$, it can only be either $(a+1)^2$ or $(a+2)^2$. Let's study these cases.
Case $(a+1)^2$
$$a^2+6b = (a+1)^2 = a^2+2a+1$$
So $a = 3b - \frac{1}{2}$ which is impossible as this isn't a natural number.
Case $(a+2)^2$
$$a^2+6b = (a+2)^2 = a^2+4a+4$$
So $a = \frac{3}{2}b - 1$. Let's plug this into the second square root and see what we get.
$$b^2 + 6a = b^2 + 6\left(\frac{3b}{2} - 1\right)
= b^2 + 9b - 6
$$
Now the square must be $(b+m)^2$ for some $m$. What can $m$ be?
If $m>4$, we have $ (b+m)^2 - (b^2 + 9b - 6) = (2m-9)b + m^2 + 6 > 0$, so $m>4$ won't work.
The case $m=4$ works and gives $b=22$.
How about $m<4$. Well, there we get the solution $b=2$.
I was a bit hasty here in the case $m<4$, since the negative values could also give solutions. But we can write $(b+m)^2 = b^2 + 9b - 6$ and get
$$b = \frac{m^2+6}{9-2m} = -\frac{m}{2} + \frac{105-9(9-2m)}{4(9-2m)}$$
so $m$ must be even and $9-2m$ a divisor of $105$. So it actually goes like in the other answer and I think that is simpler.
Answer: $a=b=2$ or $a=32, b = 22$ (or other way around).
Best Answer
While there is a solution by completing a square, I want to present another way. We have that $x^2+7x+21$ is a square of an integer.
Let $x>0$. Then $$(x+3)^2< x^2+7x+21< (x+5)^2.$$ It means that $$x^2+7x+21=(x+4)^2.$$ This gives $x= 5$. Now let $x\le-9$. Then $$(x+4)^2< x^2+7x+21< (x+3)^2+12<(x+2)^2.$$
So we have that $$x^2+7x+21=(x+3)^2.$$ This gives $x= -12$. Now we just try all $x$ from $\{-8,-7,-6,-5,-4,-3,-2,-1,0\}$ and find that $x=-3$ and $x=-4$ work, too. So the answer is $x\in\{-12,-4,-3,5\}$.