You want
$$ 3a^2 - 4 m^2 = -n^2. $$ Suppose you have a single such expression, $a=1,m=1,n=1,$ call it $(a,m) = (1,1).$ You get infinitely many such expressions, with the same fixed value of $m,$ by repeating the mapping
$$ (a,m) \mapsto (7a+8m, 6a+7m). $$
So, we have $(a,m)$ pairs
$$ (1, 1) $$
$$ (15, 13) $$
$$ (209, 181) $$
$$ (2911, 2521) $$
$$ (40545, 35113) $$
We also get separate recursions for the $(a,m)$ components,
$$ a_{j+2} = 14 a_{j+1} - a_j, $$
$$ m_{j+2} = 14 m_{j+1} - m_j. $$
This is just the Cayley-Hamilton theorem. Similarly, the mapping preserves GCD, because
$$
\left(
\begin{array}{rr}
7 & 8 \\
6 & 7
\end{array}
\right)
$$
is of integers and determinant $1,$ that is, its inverse is also of integer entries. Back to Cayley-Hamilton, what is the characteristic polynomial for this matrix?
Well, let's see. So we have
$$\begin{cases}
a^2 = x(b+c) \\
b^2 = y(a+c) \\
c^2 = z(a+b)
\end{cases}$$
where all quantities are integers. What does this mean?
For one thing, each of $a,b,c$ is strictly greater than 1.
For another, if $a,b,c$ are all pairwise coprime (as per the problem statement), then so are $b+c, a+c, a+b$.
Now let's make one peculiar combination out of these numbers, for no reason at all:
$$(a+b+c)^2 = a^2+2ab+2ac+(b+c)^2=x(b+c)+2a(b+c)+(b+c)^2$$
So it is divisible by $b+c$, and also (following similar reasoning) by $a+b$ and $a+c$. If these three are all mutually coprime, then it is divisible by their product.
Assume that $a<b<c$. Then $b+c>{2\over3}(a+b+c)$ and $a+c>{1\over3}(a+b+c)$. So...
$$(a+b)(b+c)(a+c)\;|\;(a+b+c)^2\\
(a+b)(b+c)(a+c)\leqslant(a+b+c)^2\\
(a+b)\cdot{1\over3}(a+b+c)\cdot{2\over3}(a+b+c)<(a+b+c)^2\\
a+b < {9\over2}\\
a+b \leqslant 4$$
With $a,b>1$ this only leaves us the option of $a=b=2$, which are not coprime.
Finally, there are no such coprime triples. Q.e.d.
Best Answer
Hint. Show the infinitude of positive integer solutions $(a,b)$ to the divisibility condition $ab\mid a^2+b^2-5$. In fact, for a positive integer $k$, there exists $(a,b)\in\mathbb{Z}_{>0}\times\mathbb{Z}_{>0}$ such that $$a^2+b^2-5=kab\tag{*}$$ if and only if $k=3$, in which case there are infinitely many choices of $(a,b)$. When $k=3$, amongst the positive integer solutions $(a,b)$ such that $a\geq b$, the smallest of which is $(a,b)=(4,1)$.
The idea is the technique known as Vieta jumping. If you do this correctly, then you will see that all positive integer solutions $(a,b)$ with $a\geq b$ to (*) with $k=3$ are of the form $(a,b)=(x_n,x_{n-1})$ for some positive integer $n$, where $(x_n)_{n=0}^\infty$ are given by $x_0=1$, $x_1=4$, and $$x_n=3x_{n-1}-x_{n-2}$$ for every integer $n\geq 2$. Here is a closed form of $\left(x_n\right)_{n=0}^\infty$: $$x_n=\left(\frac{1+\sqrt5}{2}\right)^{2n+1}+\left(\frac{1-\sqrt5}{2}\right)^{2n+1}=L_{2n+1}$$ for all $n=0,1,2,\ldots$, where $(L_r)_{r=0}^\infty$ is the sequence of Lucas numbers. The first few terms of $(x_n)_{n=0}^\infty$ are $$1,4,11,29,76,199,521,1364,3571,9349,24476,\ldots\,.$$ Compare the list above with the answer by Arthur.