(This might need some editing later, it's a bit messy.)
As a summary, for any given $a$ there are always solutions for the equation
$$
\varphi(a^2) + \varphi(b^2) = \varphi(c^2)
$$
The approach is similar to the one described in the OP. We consider all cases of prime factors $2,3,5,7$ of $a$, then find appropriate $b$ and $c$ using that information.
Let $r,s,t$ be the exponent of prime factors $2,3,5$ of $a$. We write
$$
a=2^r3^s5^tm
$$
Proposition: If $(r,s,t)$ is not of the form $r=1,s=1$ and $t\geq 2$, then there is a solution.
We shall solve the various cases as follows:
Case 1: $r=0$.
Case 2: $r\geq 1$ and $s=0$.
Case 3: All remaining cases except $r=1,s=1$ and $t\geq 2$.
For the exceptional case $r=1,s=1,t\geq 2$, we require prime $7$. This is handled in case 4.
Case 1: $r=0$
This case is covered in the OP: Since $\gcd(r,2)=1$,
$$
\varphi(a^2)+\varphi(a^2)=2\varphi(a^2)= \varphi((2a)^2)
$$
so we set $b=a,c=2a$.
Hence we assume $r\geq 1$.
For case 2 we cover the cases $s=0$.
Case 2a: $r=1,s=0$ any $t$
We have
$$
\varphi(2^2) + \varphi(3^2) = 8 = \varphi((2^2)^2)
$$
therefore
$$
\varphi((2\cdot 5^tm)^2) + \varphi((3\cdot 5^tm)^2) = \varphi((2^2\cdot 5^tm)^2)
$$
Case 2b: $r\geq 2,s=0,t=0$
Let
$$
b = 2^{r+1}m ,\quad c = 2^{r-1}5m
$$
Then
$$
\begin{align}
\varphi(a^2)+\varphi(b^2) &= \varphi((2^r m)^2) + \varphi((2^{r+1} m)^2)\\
&= 2^{2r-1}\varphi(m^2) + 2^{2r+1} \varphi(m^2) \\
&= 2^{2r-1}5 \varphi(m^2)\\
& = \varphi(2^{2r-2}\cdot 5^2m^2)\\
&= \varphi(c^2)
\end{align}
$$
Notice we needed $r\geq 2$ for equality 4.
Case 2c: $r\geq 2,s=0,t\geq 1$
Let
$$
b = 2^{r+1}3^15^t m, \quad c = 2^r 5^{t+1}m
$$
which gives
$$
\begin{align}
\varphi(a^2) + \varphi(b^2) &= \varphi(2^{2r}5^{2t}m^2) + \varphi(2^{2r+2} 3^2 5^{2t} m^2)\\
&= 2^{2r+1}5^{2t-1} \varphi(m^2) + 2^{2r+4}\cdot 3\cdot 5^{2t-1} \varphi(m^2)\\
&= (1 + 8\cdot 3)2^{2r+1}5^{2t-1}\varphi(m^2)\\
&= 2^{2r+1} 5^{2t+1} \varphi(m^2)\\
&= \varphi(2^{2r} 5^{2t+2} m^2)\\
&= \varphi(c^2)
\end{align}
$$
Observe that this would have worked with $r=1$ as well.
Hence we may now consider $r,s\geq 1$.
Solve the case of $t=0$ first:
Case 3a: $r,s\geq 1, t=0$
Let
$$
b = 2^{r+1}3^s5^1 m,\quad c= 2^r 3^{s+2}m
$$
and we verify that
$$
\begin{align}
\varphi(a^2) + \varphi(b^2) &= \varphi(2^{2r}3^{2s}m^2) + \varphi(2^{2r+2}3^{2s}5^2m^2)\\
&= 2^{2r}3^{2s-1}\varphi(m^2) + 2^{2r+4}3^{2s-1}5\varphi(m^2)\\
&= (1 + 2^4\cdot 5)2^{2r}3^{2s-1}\varphi(m^2)\\
&= 2^{2r} 3^{2s+3} \varphi(m^2)\\
&= \varphi(2^{2r} 3^{2s+4}m^2)\\
&= \varphi(c^2)
\end{align}
$$
This leaves the case of $r,s,t\geq 1$. We first eliminate the cases of $r\geq 3$.
Case 3b: $r\geq 3,s,t\geq 1$
This can be seen directly from
$$
\begin{align}
\varphi((2^{r}3^{s}5^{t}m)^2) + \varphi((2^{r-2}3^{s+1} 5^t m)^2 &= \varphi(2^{2r} 3^{2s} 5^{2t} m^2) + \varphi(2^{2r-4} 3^{2s+2} 5^{2t} m^2)\\
&= 2^{2r+2} 3^{2s-1} 5^{2t-1} \varphi(m^2) + 2^{2r-2} 3^{2s+1} 5^{2t-1} \varphi(m^2)\\
&= (2^4 + 3^2)2^{2r-2}3^{2s-1} 5^{2t-1} \varphi(m^2)\\
&= 2^{2r-2}3^{2s-1} 5^{2t+1} \varphi(m^2)\\
&= \varphi( 2^{2r-4} 3^{2s} 5^{2t+2} m^2)\\
&= \varphi( (2^{r-2} 3^s 5^{t+1} m)^2 )
\end{align}
$$
The condition $r\geq 3$ is used in equality 2, for computing exponents of prime $2$.
We are left with the cases $r\in \{1,2\}$ and $s,t\geq 1$.
Case 3c: $r=2$ and $s,t\geq 1$
Let
$$
b = 3^s 5^t m,\quad c = 3^{s+1} 5^t m
$$
Then
$$
\begin{align}
\varphi(a^2) + \varphi(b^2) &= \varphi(2^4 3^{2s} 5^{2t}m^2) + \varphi(3^{2s} 5^{2t} m^2)\\
&= 2^6 3^{2s-1} 5^{2t-1} \varphi(m^2) + 2^3 3^{2s-1} 5^{2t-1} \varphi(m^2)\\
&= (2^3+1) 2^3 3^{2s-1}5^{2t-1} \varphi(m^2)\\
&= 2^3 3^{2s+1} 5^{2t-1} \varphi(m^2)\\
&= \varphi( 3^{2s+2} 5^{2t} m^2)\\
&= \varphi(c^2)
\end{align}
$$
Finally, we investigate the case of $r=1$. When $s\geq 2$ we still have a solution:
Case 3d: $r=1, s\geq 2$ and $t\geq 1$
Let
$$
b = 2^{3} 3^{s-1} 5^t m, \quad c = 2^1 3^{s-1} 5^{t+1} m
$$
and we check that
$$
\begin{align}
\varphi(a^2) + \varphi(b^2) &= \varphi( 2^{2} 3^{2s} 5^{2t} m^2) + \varphi( 2^{6} 3^{2s-2} 5^{2t} m^2)\\
&= 2^{4} 3^{2s-1} 5^{2t-1} \varphi(m^2) + 2^{8} 3^{2s-3} 5^{2t-1} \varphi(m^2)\\
&= (3^2 + 2^4)2^{4} 3^{2s-3} 5^{2t-1} \varphi(m^2)\\
&= 2^{4} 3^{2s-3} 5^{2t+1} \varphi(m^2)\\
&= \varphi( 2^{2} 3^{2s-2} 5^{2t+2} m^2 )\\
&= \varphi(c^2)
\end{align}
$$
However in the final case of $r=1,s=1$, we only have solution for $t=1$
Case 3e: $r=1,s=1,t=1$
This can be seen directly from
$$
\begin{align}
\varphi((2^1 3^s 5^1 m)^2) + \varphi( (2^3 3^s m)^2 ) &= \varphi( 2^2 3^{2s} 5^{2} m^2) + \varphi( 2^6 3^{2s} m^2)\\
&= 2^4 3^{2s-1} 5^1 \varphi(m^2) + 2^6 3^{2s-1} \varphi(m^2)\\
&= (5 + 2^2)2^4 3^{2s-1} \varphi(m^2)\\
&= 2^4 3^{2s+1} \varphi(m^2)\\
&= \varphi( 2^4 3^{2s+2} m^2) \\
&= \varphi( (2^2 3^{s+1} m)^2 )
\end{align}
$$
The unsolved case is $r=1,s=1$ and $t\geq 2$, which we handle next.
Case 4: $r=1,s=1,t\geq 2$
Let $u$ be the exponent of the prime factor $7$ of $a$. i.e.
$$
a = 2^1 3^1 5^t 7^u n
$$
If $u=0$, we set
$$
b = 5^t 7^1 n,\quad c = 3^2 5^t n
$$
so that
$$
\begin{align}
\varphi(a^2)+\varphi(b^2) &= 48\cdot 5^{2t-1}\varphi(n^2) + 168\cdot 5^{2t-1} \varphi(n^2)\\
&= 2\cdot 3^3 \cdot 4\cdot 5^{2t-1} \varphi(n^2)\\
&= \varphi(3^4 5^{2t} n^2)\\
&= \varphi(c^2)
\end{align}
$$
For $u=1$, we set
$$
b = 2^2 5^t n,\quad c = 2^5 5^t n
$$
Checking:
$$
\begin{align}
\varphi(a^2) + \varphi(b^2) &= \varphi(2^2 3^2 5^{2t} 7^2 n^2) + \varphi( 2^4 5^{2t} n^2)\\
&= 2^5 3^2 5^{2t-1} 7^1 \varphi(n^2) + 2^5 5^{2t-1}\varphi( n^2)\\
&= (3^2 7^1 + 1) 2^5 5^{2t-1} \varphi(n^2)\\
&= 2^{11} 5^{2t-1} \varphi(n^2)\\
&= \varphi( 2^{10} 5^{2t} n^2)\\
&= \varphi(c^2)
\end{align}
$$
Finally, for the last case of $u\geq 2$, we can set
$$
b = 2^4 3^2 5^t 7^{u-1} n,\quad c = 2^1 3^1 5^{t+2} 7^{u-1} n
$$
Checking:
$$
\begin{align}
\varphi(a^2) + \varphi(b^2) &= \varphi(2^2 3^2 5^{2t} 7^{2u} n^2) + \varphi( 2^8 3^4 5^{2t} 7^{2u-2} n^2)\\
&= 2^5 3^2 5^{2t-1} 7^{2u-1} \varphi(n^2) + 2^{11} 3^{4} 5^{2t-1} 7^{2u-3} \varphi(n^2)\\
&= (7^2 + 2^6 3^2) 2^5 3^2 5^{2t-1} 7^{2u-3} \varphi(n^2)\\
&= (625) 2^5 3^2 5^{2t-1} 7^{2u-3} \varphi(n^2)\\
&= 2^5 3^2 5^{2t+3} 7^{2u-3} \varphi(n^2)\\
&= \varphi( 2^2 3^2 5^{2t+4} 7^{2u-2} n^2)\\
&= \varphi(c^2)
\end{align}
$$
Best Answer
Regarding your method, you limit the primes of interest nicely, but you also need to eliminate from consideration any of those primes which do not pass the basic divisibility test of $\phi(p_i) \mid 14$. This would leave you with only $2$ and $3$ which clearly cannot produce the factor of $7$ needed to make $\phi(n)=14$.
Thinking about this problem separately from your approach, I would start with the key aspects:
a) $14+1 = 15$ is not a prime, and
b) $14=2\cdot 7$
So from (a) we do not arrive at $\phi(n)=14$ directly, and from (b) we need a factor of $7$. However that can only be supplied by a higher power of $7$ - which cannot be present as this would also include a factor of $6$ - or a larger prime (e.g. $29$), which would also include a larger factor not present.
So there is no number $n$ with $\phi(n)=14$.
The OEIS A005277 sequence gives even non-totients; $14$ is the smallest.