I have found references to a proof that any odd triperfect is a square.
See here and here. They both reference the following German paper:
H.-J. Kanold, "Über mehrfach vollkommene Zahlen. II," J. Reine Angew. Math., v. 197, 1957, pp. 82-96. MR 18, 873.
I'm in the process of translating and extracting the relevant portion and will edit that in soon.
Update:
From p. 88-89, here's what I have extracted:
Lemma 1.
Let $\displaystyle n = \prod_{i=1}^k p_i^{\alpha_i}$ be an $(s-1)$-fold perfect number and also let $sn \equiv 1 \pmod 2$. Then $n > 10^{20}$.
Proof: We start from the relationship
$$sn = s \prod_{i=1}^k p_i^{\alpha_i} = \prod_{i=1}^k \sigma\bigl(p_i^{\alpha_i}\bigr).$$
From this, we see that
$$\alpha_i \equiv 0 \pmod 2 \text{ for } i = 1, \ldots, k$$
must be true.
[Remainder of proof excluded as this is all we need.]
This is what we want since $sn \equiv 1 \pmod 2$ means $n$ is odd (since $sn$ is odd implies both $s$ and $n$ are odd) and $\alpha_i \equiv 0 \pmod 2$ means that $n$ is a perfect square (since each prime power is even).
To explain this a bit more, suppose some $\alpha_j$ is odd and use the following formula (reference):
$$\sigma\bigl(p_j^{\alpha_j}\bigr) = 1 + p_j + p_j^2 + \cdots + p_j^{\alpha_j}.$$
Since $n$ is odd, we know $p_j$ must be odd and hence any power of $p_j$ is also odd. Thus we have an odd number of odd integers plus the remaining $1$. Thus the sum is even, which implies $sn$ is even. This is a contradiction. Therefore, all the $\alpha_i$ must be even and hence $n$ is a perfect square.
Edit: (Clarification of $(s-1)$-fold perfect number notation)
This is the first couple sentences from the paper:
Wir schließen uns in dieser Arbeit der Bezeichnungsweise einer früheren an Danach heißt eine naturliche Zahl $\displaystyle n=\prod_{x=1}^k p_x^{\alpha_x}$ eine $(s-1)$-fach vollkommene Zahl, wenn sie der Bedingung $\sigma(n) = s \cdot n$ genügt. Die $p_x$ bedeuten Primzahlen, $\sigma(n)$ bezeichnet die Summe aller positiven Teiler von $n$.
I have translated this as:
We agree in this work with earlier notation used before by calling a natural number $\displaystyle n=\prod_{x=1}^k p_x^{\alpha_x}$ an $(s-1)$-fold perfect number if satisfies the condition $\sigma(n) = s \cdot n$. The $p_x$ denote primes and $\sigma(n)$ denotes the sum of all positive divisors of $n$.
This means what they refer to as a "$(s-1)$-fold perfect number" is what we would call an "$s$-perfect number". This understanding also agrees with the proof.
